Analytic Geometry
Straight Line
Learning Objectives:
Find and interpret equation of a straight line in various forms.
Perform slope calculations including parallel and perpendicular lines.
Find angle between two coplanar, nonparallel lines.
The equation of a straight line (or any curve) is the relation between the x and y (and z) coordinates of all points lying on it.
The general form of the equation of a straight line is: \[Ax + By + C = 0\]
Slope of a straight line gives you an idea about its inclination with reference to xaxis. Slope is also referred as gradient.
Slope of a Straight Line:
\[\mathrm{Slope \ of \ a \ straight \ line} = \dfrac{\mathrm{Rise}}{\mathrm{Run}} = \dfrac{\Delta y}{\Delta x}\]
\[m = \dfrac{y_{2}y_{1}}{x_{2}x_{1}}\]
Equation of a Straight Line:
Slope Intercept Format:
\[y = mx + b\]
where m = slope and b = yintercept
For the above line, yintercept = 1, and
slope = \(\dfrac{\mathrm{Rise}}{\mathrm{Run}}\) = \(\dfrac{2}{4}\) = \(\dfrac{1}{2}\)
So the equation will be, \[y = 0.5x + 1\] \[2y = x + 2\] \[x  2y + 2 = 0\]
Slope Point Format:
\[y  y_{1} = m (x  x_{1})\]
where \(x_{1}\) and \(y_{1}\) are the coordinates of the point through which the line passes.
Double Intercept Format:
\[\dfrac{x}{a}+\dfrac{y}{b} = 1\]
where a = x intercept and b = y intercept
For parallel lines slopes are equal i.e. \(m_{1} = m_{2}\)
For perpendicular lines
\(m_{1} * m_{2}\) = 1
or,
\[m_2 = \dfrac{1}{m_1}\]
Angle between Two Straight Lines:
Angle between two straight lines is given by: \[\alpha = \tan^{1}\left(\dfrac{m_{2}m_{1}}{1+ m_{1}m_{2}}\right)\]
Solved Example: 1101
The angle between lines 2x 9y + 16=0 and x + 4y + 5 = 0 is given by:
A. $\tan^{1}\left( \dfrac{2}{9}\right)$
B. $\tan^{1}\left( \dfrac{1}{4} \right)$
C. $\tan^{1}\left( \dfrac{1}{2} \right)$
D. The lines are parallel.
First let's find out slopes of each line. \[2x 9y + 16 =0 \mathrm{\ Or, \ } y = \dfrac{2}{9}x+\dfrac{16}{9}\] $\mathrm{\ which\ means\ slope\ } m_1 =\dfrac{2}{9}$ Similarly, slope of the second line, $m_2 = \dfrac{1}{4}$ The angle between two lines is given by: \begin{align*} \alpha &= \tan^{1}\left(\dfrac{m_{2}m_{1}}{1+ m_{1}m_{2}}\right )\\ &= \tan^{1}\left(\dfrac{\dfrac{1}{4}\dfrac{2}{9}}{1+ \dfrac{2}{9}\times \dfrac{1}{4}}\right )\\ &=\tan^{1} \left( {\dfrac{ \dfrac{17}{36}}{\dfrac{34}{36}}}\right )\\ &=\tan^{1}\left(\dfrac{1}{2}\right) \end{align*}
Correct Answer: C
Solved Example: 1102
Consider the point $P (1,2)$ and the line L: $x  2y + 2 = 0$. Which of the following is a correct statement?
A. Point P lies on the line L.
B. Point P does not lie on the line L and it is located towards the origin side.
C. Point P does not lie on the line L and it is located towards the nonorigin side.
D. Point P does not lie on the line L but it impossible to find which side it lies unless additional information is given.
The line L: $x  2y + 2 = 0$ can be brought to slopeintercept format as: $y =0.5x + 1$
A rough sketch can be drawn beginning yintercept point (0,1).
The line has a positive slope, means uphill (from left to right).
The slope is 0.5 or $\dfrac{1}{2}$ means every 2 units it travels horizontally (run), it travels 1 unit along yaxis (rise).
From this information, it can be found out that the point P is towards the nonorigin side.Alternate method:
Substitute (0,0) in the given equation of L. We get 00+2 = 2>0 for Origin.
Substitute (1,2) in the given equation of L. We get 14+2 = 1<0 for (1,2).
Since both answers have opposite sign, the point P is on the opposite side of the origin.
Correct Answer: C
Solved Example: 1103
Given the slopeintercept form of a line as y = mx + b, which one of the following is true?
A. $\dfrac{m}{b}$ is the y axis intercept
B. b is the x axis intercept
C. y  b is the slope of the line
D. $\dfrac{b}{m}$ is the x axis intercept
For the straight line y = mx + b, m is the slope and b is the yintercept. To get xintercept, we will have to substitute y=0: $y = mx + b, 0 = mx + b, \dfrac{b}{m} = x$
Correct Answer: D
Solved Example: 1104
A ray of light coming from the point A(1, 2) is reflected at a point P(x,0) on the xaxis and then passes through the point B(5, 3). The coordinates of the point A are: (BITSAT 2016)
A. ($\dfrac{13}{5}$,0)
B. ($\dfrac{5}{13}$,0)
C. ( 7, 0)
D. ( $\dfrac{9}{2}$, 0)
From similarity of triangles, \[\dfrac{AM}{MP} = \dfrac{BN}{NP}\] \[ \dfrac{2}{x1} = \dfrac{3}{5x}, x = \dfrac{13}{5}\]Correct Answer: A
Solved Example: 1105
If the coordinates of the middle point of the portion of a line intercepted between coordinate axes (3,2), then the equation of the line will be:
A. 2x+3y=12
B. 3x+2y=12
C. 4x3y=6
D. 5x2y=10
As shown in the figure, the coordinates of intersection point of the straight line with x axis will be:B (2$\times$3, 0)= B (6,0)
and yaxis will be:A(0, 2$\times$2) = A(0,4).
So, xintercept = a = 6, yintercept = b= 4Using double intercept form of straight line equation,
$\dfrac{x}{a} + \dfrac{y}{b} = 1, \dfrac{x}{6} + \dfrac{y}{4} = 1, 2x+3y =12$
Correct Answer: A
Solved Example: 1106
The image of the point (4, 3) with respect to the line y = x is:
A. (4, 3)
B. (3, 4)
C. (4, 3)
D. (3, 4)
Slope of original (red) line = m$_1$ = 1
Since reflection (blue) line is perpendicular to original (red) line,Slope of reflection (blue) line m$_2$ = $\dfrac{1}{m_1}$ = 1
Let M(a,a) be the point of intersection as shown in the figure. We have taken both coordinates same for M, because it also lies on y=x line.Now slope of MA = 1
\[\dfrac{3a}{4a} = 1, a= 0.5\] But M is the midpoint of AB. Using midpoint formula, \[\dfrac{p+4}{2} = 0.5, \mathrm{\ or,\ } p = 3 \] \[\mathrm{similarly\ }\dfrac{q3}{2} = 0.5, \mathrm{\ or,\ } q = 4 \mathrm{\ or,\ } B = (3,4)\]
Correct Answer: D
Solved Example: 1107
A line L passes thro' the points (1, 1) and (2, 0) and another line $L_2$ passes through ($\dfrac{1}{2}$,0) and $\perp$ to $L$. Then the area of the triangle formed by the lines $L$, $L_2$ and yaxis, is:
A. $\dfrac{15}{8}$
B. $\dfrac{25}{4}$
C. $\dfrac{25}{8}$
D. $\dfrac{25}{16}$
Slope of first (red) line = m$_1$ = $\dfrac{y_2  y_1}{x_2  x_1}$ = $\dfrac{01}{21}$ = 1
Equation of first (red) line:\[y  y_1 = m(x x_1)\] \[y 0 = 1 (x  2)\] \[ y = x + 2\]
Slope of second (blue) line = $\dfrac{1}{m_1}$ = 1
Equation of second (blue) line: $y  y_1 = m(x x_1), y 0 = 1 (x  0.5), y = x 0.5$Solving these two equations simultaneously, we will get the intersection point as M (1.25,0.75).
\begin{align*} \mathrm{Area\ of\ Triangle} &=\dfrac{1}{2} \times \mathrm{base} \times \mathrm{height}\\ &=\dfrac{1}{2} \times 2.5 \times 1.25\\ &= \dfrac{25}{16} \end{align*}
Correct Answer: D
Solved Example: 1108
The area of triangle formed by the lines $x =0, y =0\ \mathrm{and} \dfrac{x}{a} + \dfrac{y}{b} =1$, is:
A. ab
B. $\dfrac{ab}{2}$
C. 2ab
D. $\dfrac{ab}{3}$
The equation of the line $\dfrac{x}{a} + \dfrac{y}{b} =1$ is in double intercept form. This line makes an intercept 'a' on the xaxis and 'b' on the yaxis.
Area of the right angled triangle is = $\dfrac{1}{2}$ $\times$ (base) $\times$ (Height) = $\dfrac{1}{2}$ $\times$ a $\times$ b.
Correct Answer: B
Solved Example: 1109
The diagonals of a parallelogram PQRS are along the lines x+3y=4 and 6x2y=7. Then PQRS must be a:
A. Rectangle
B. Square
C. Cyclic quadrilateral
D. Rhombus
$m_1=\dfrac{1}{3}$ and $m_2=3$. Hence lines x +3y =4 and 6x 2y =7 are perpendicular to each other. Therefore the parallelogram is rhombus.
Correct Answer: B
Solved Example: 1110
The area enclosed within the curve x+y=1 is:
A. $\sqrt{2}$
B. 1
C. $\sqrt{3}$
D. 2
In quadrant 1 (Q1), since both x and y are positive, x = x, y = y and the line will be: x + y = 1
In quadrant 2 (Q2), since x is negative and y is positive, x = x, y = y the line will be: x + y = 1
In quadrant 3 (Q3), since both x and y are negative, x = x, y = y the line will be: x  y = 1
In quadrant 4 (Q4), since x is positive and y is negative, x = x, y = y the line will be: x  y = 1
Required area = 4 \(\times\) Area in the first quadrant = 4 \(\times \dfrac{1}{2} \times\) base \(\times\) height = 4 \(\times \dfrac{1}{2} \times\) 1 \(\times\) 1 = 2
Correct Answer: D
Solved Example: 1111
If the equation of base of an equilateral triangle is 2xy=1 and the vertex is (1, 2), then the length of the side of the triangle is:
A. $\sqrt{\dfrac{20}{3}}$
B. $\dfrac{2}{\sqrt{15}}$
C. $\sqrt{\dfrac{8}{15}}$
D. $\sqrt{\dfrac{15}{2}}$
\begin{align*} \tan 60^\circ &=\dfrac {h}{\left( \dfrac {x}{2}\right) } \sqrt {3}=\dfrac {h}{\left(\dfrac {x}{2}\right)}\\ h&=\dfrac {\sqrt {3}}{2}x, \mathrm{or},\ x =\dfrac {2}{\sqrt {3}}h \end{align*} \begin{align*} \mathrm{Distance}&=\left\vert \dfrac {2\left( x_1\right) \left( y_1\right) 1}{\sqrt {2^{2}+\left( 1\right) ^{2}}}\right\vert\\ &=\left \dfrac {2\left( 1\right) \left( 2\right) 1}{\sqrt {2^{2}+\left( 1\right) ^{2}}}\right\\ &=\left \dfrac {5}{\sqrt {5}}\right =\sqrt {5} \end{align*} \begin{align*} x &=\dfrac {2}{\sqrt{3}}h\\ &=\dfrac {2}{\sqrt{3}}\sqrt{5}\\ &=\dfrac {\sqrt {4}\sqrt {5}}{\sqrt{3}}\\ &=\sqrt {\dfrac {20}{3}} \end{align*}Correct Answer: A
Solved Example: 1112
Let PS be the median of the triangle with vertices P (2,2), Q (6,1) and R (7,3). The equation of the line passing through (1,1) and parallel to PS is:
A. 2x9y7=0
B. 2x9y11=0
C. 2x+9y11=0
D. 2x+9y+7=0
S = midpoint of QR=($\dfrac{6+7}{2}$,$\dfrac{1+3}{2}$)=($\dfrac{13}{2}$,1) $PS=\dfrac{21}{2\dfrac{13}{2}}=\dfrac{2}{9},$
The required equation is: $y+1 =\dfrac{2}{9}(x1), \mathrm{Or,\ }2x+9y+7 =0$.
Correct Answer: D
Solved Example: 1113
The area bounded by the curves $x + 2 \vert y \vert = 1 \mathrm{\ and\ } x = 0$ is:
A. $\dfrac{1}{4}$
B. $\dfrac{1}{2}$
C. 1
D. 2
When y < 0, $\vert$y$\vert$ = y, So when y < 0, the equation becomes, x  2y = 1
When y > 0, $\vert$y$\vert$ = +y, So when y > 0,the equation becomes, x + 2y = 1
Area of Triangle,$A = \dfrac{1}{2} \times b \times h = \dfrac{1}{2} \times 1 \times 1 = \dfrac{1}{2} \mathrm{\ sq.\ units}$
Correct Answer: B
Conics
Learning Objectives:
Understand origin of a conic section as a section of cone by a cutting plane at various angles.
Recognize the conic sections from their functions in standard from and from their graphs.

If B\(^2\)  4AC <0, an ellipse is defined.

If B\(^2\)  4AC >0, a hyperbola is defined.

If B\(^2\)  4AC = 0, the conic is a parabola.

If A = C and B = 0, a circle is defined.

If A = B = C = 0, a straight line is defined.
\[x^2 + y^2 + 2ax + 2by + c = 0\] is the normal form of the conic section equation, if that conic section has a principal axis parallel to a coordinate axis. \[h = –a; k = –b\]
\[r = \sqrt{a^2 + b^2  c}\]

If a\(^2\) + b\(^2\)  c is positive, a circle, center (a, b).

If a\(^2\) + b\(^2\)  c equals zero, a point at (a, b).

If a\(^2\) + b\(^2\)  c is negative, locus is imaginary
Solved Example: 1201
6x$^2$ + 12x + 6y$^2$  8y =100 is an example of a:
A. Circle
B. Parabola
C. Hyperbola
D. Ellipse
\begin{align*} 6x^{2}+12x+6y^{2}8y &=100\\ \left[ 6x^{2}+12x\right] +\left[ 6y^{2}8y\right] &=100\\ 6\left[ x^{2}+2x+1\right] +6\left[ y^{2}\dfrac {4}{3}y+\dfrac {4}{9}\right] &= 100+ 6 + \dfrac {24}{9}\\ 6\left[ x+1\right]^{2}+6\left[ y\dfrac {2}{3}\right]^{2}&= 106 + \dfrac {24}{9}\\ \left( x+1\right)^{2}+\left( y\dfrac {2}{3}\right)^{2} &= \dfrac {978}{6} \end{align*} It is a circle, with center = (1,$\dfrac{2}{3}$)
Correct Answer: A
Solved Example: 1202
4x$^2$ + 12x + y$^2$  8y = 64 is an example of a:
A. Circle
B. Parabola
C. Hyperbola
D. Ellipse
\begin{align*} 4x^{2}+12x+y^{2}8y&=64\\ \left[ 4x^{2}+12x\right] +\left[ y^{2}8y\right] &=64\\ 4\left[ x^{2}+3x+\dfrac {9}{4}\right] +\left[ y^{2}8y+16\right]&= 64 + 9 + 16\\ 4\left[ x+\dfrac {3}{2}\right] ^{2}+\left[ y4\right] ^{2}&=89 \end{align*} It is an ellipse.
Correct Answer: D
Solved Example: 1203
6y = 3x$^2$  15 is an example of a:
A. Circle
B. Parabola
C. Hyperbola
D. Ellipse
\begin{align*} 6y &=3x^{2}15\\ 3x^{2}&=6y+15\\ x^{2}&=2y+5\\ x^{2}&=2\left[ y+\dfrac {5}{2}\right]\\ X^2 &= 2Y \end{align*} It is a parabola.Correct Answer: B
Parabola
Learning Objectives:

Find equation of parabola Types of parabolas, eccentricity directrix, focus, latus rectum, asymptote.
The general equation of parabola is: \[(y  k)^2 = 2p(x  h)\] with center at (h, k).
The standard form of a parabola’s equation is, when h = k = 0, and it is generally expressed as: \[y^2 = 4ax\] With focus: (\(\dfrac{p}{2}\), 0) and directrix: x = \(–\dfrac{p}{2}\)
If a >0, the parabola opens towards right, if a <0, it opens towards left.
\[x^2 = 4by\] If b >0, the parabola opens upwards, if b <0, it opens downwards.
Key Properties of a parabola:

Vertex is the minimum / maximum point of the parabola.

Focus is a point on the axis of symmetry of the parabola that is a set distance from the vertex of the parabola.

Directrix is a line (not a point) that is perpendicular to the axis of symmetry of the parabola and does not intersect with the parabola.

Eccentricity of a parabola is always equal to 1 because any point on parabola is equidistant from focus and directrix.
Solved Example: 1501
3$x^2$ + 2x  5y + 7 = 0. Determine the curve.
A. Parabola
B. Ellipse
C. Hyperbola
D. Circle
By rearranging the terms and completing square, we get $(\sqrt{3}x + \dfrac{1}{\sqrt{3}})^2 = 5 (y  \dfrac{4}{3})$
which can be represented as, $X^2 = 4bY$
Hence, it is a parabola vertically facing upwards.
Correct Answer: A
Solved Example: 1502
The focus of the parabola $y^2$ = 16x is at:
A. (4, 0)
B. (0, 4)
C. (3, 0)
D. (0, 3)
Comparing with the standard equation of parabola, $y^2 = 4ax$ we get a = 4. The focus is at (a,0) which will be at (4,0).
Correct Answer: A
Solved Example: 1503
Where is the vertex of the parabola $x^2$ = 4(y  2)?
A. (2, 0)
B. (0, 2)
C. (3, 0)
D. (0, 3)
This is a vertically upward facing parabola which is shifted upwards from the origin along yaxis by 2 units. Hence, the vertex will be (0,2).
Correct Answer: B
Solved Example: 1504
Find the equation of the directrix of the parabola $y^2$ = 16x.
A. x = 2
B. x = 2
C. x = 4
D. x = 4
Comparing $y^2$ = 16x with $y^2$ = 4ax, we get, a= 4. The equation of the directrix is x = a, Hence, x = 4.
Correct Answer: D
Solved Example: 1505
Given the equation of a parabola $3x + 2y^2 4y + 7 = 0$. Locate its vertex.
A. $(\dfrac{5}{3}, 1)$
B. $(\dfrac{5}{3}, 1)$
C. $(\dfrac{5}{3}, 1)$
D. $(\dfrac{5}{3}, 1)$
\[3x+2y^{2}4y+7=0, \mathrm{\ or\ } \left(2y^{2}4y \right) +7=3x\]
Completing a perfect square term from the first two terms,
last term is given by:$=\dfrac {\left(\mathrm{middle\ term}\right) ^{2}}{4\left(\mathrm{first\ term}\right) } =\dfrac {\left( 4\right) ^{2}}{4\left( 2\right) }=\dfrac {16}{8}=2$
\[2y^{2}4y+7+2=3x+2\]
\[\left( \sqrt {2}y\sqrt {2}\right) ^{2}+7=3x+2\]
\[\left( \sqrt {2}y\sqrt {2}\right) ^{2}=3x5\]
\[\left( y1\right) ^{2}=\dfrac {3}{\sqrt {2}}\left[ x+\dfrac {5}{3}\right]\]
Comparing with $Y^{2}=4aX$ where $ Y = (y1)$ and $X = x + \dfrac{5}{3}$
Take (opposite of term with x, opposite of term with y). vertex = $\left( \dfrac {5}{3},1\right) $
Correct Answer: D
Solved Example: 1506
In the equation $y =  x^2 + x + 1$ where is the curve facing?
A. Downward
B. Facing left
C. Facing right
D. Upward
\begin{align*} y &=x^{2}+x+1 \\ &=\left[ x^{2}x1\right] \\ &=\left[ \left( x\dfrac {1}{2}\right) ^{2}\dfrac {5}{4}\right] \end{align*} \[y\dfrac {5}{4}=\left( x\dfrac {1}{2}\right) ^{2}\] Comparing with: $Y=X^{2}$ where $X=x\dfrac {1}{2},$ $Y=y\dfrac {5}{4}$Correct Answer: A
Solved Example: 1507
Find the location of the focus of the parabola $y^2 + 4x  4y  8 = 0$
A. (2.5, 2)
B. (3, 1)
C. (2, 2)
D. (2.5, 2)
\begin{align*} y^{2}4y &=4x+8\\ y^{2}4y+4 &=4x+8+4\\ \left( y2\right) ^{2} &=4\left( x3\right)\\ Y^{2} &=4aX\\ X &=x3\\ Y &=y2\\ a &=1 \end{align*} \[\mathrm{Vertex}=\left( 3,2\right)\] This parabola is facing left. Focus will be 'a' units towards left. Focus = (2,2)
Correct Answer: C
Solved Example: 1508
Find the equation of the axis of symmetry of the function $y = 2x^2  7x + 5$
A. 7x + 4 = 0
B. 4x + 7 = 0
C. 4x  7 = 0
D. x  2 = 0
\[y = 2x^2  7x + 5\] \[y = 2(x^2  \dfrac{7}{2}x) + 5\] \[y = 2(x^2  \dfrac{7}{2}x+ \dfrac{49}{16})\dfrac{49}{8} + 5\] \[y=2(x\dfrac{7}{4})^2\dfrac{9}{8}\] \[(y+\dfrac{9}{8}) =2(x\dfrac{7}{4})^2\] \[(x\dfrac{7}{4})^2 = 4 \times \dfrac{1}{8}(y+\dfrac{9}{8})\] \[X^2 = 4 \times \dfrac{1}{8} Y\] The equation of the axis of symmetry will be, $X = 0. x\dfrac{7}{4}= 0, x=\dfrac{7}{4}, 4x  7 = 0$Correct Answer: C
Solved Example: 1509
A parabola has its focus at (7, 4) and directrix y = 2. Find its equation.
A. $x^2$ + 12y  14x + 61 = 0
B. $x^2$  14y + 12x + 61 = 0
C. $x^2$  12x + 14y + 61 = 0
D. None of the above
Let P(x,y) by a point on the required parabola. From the focusdirectrix property, the distance from the focus and the distance from directrix must be the same. \[(x7)^2 + (y+4)^2 = (y2)^2\] \[x^2  14x + 12y + 61 = 0\]
Correct Answer: A
Solved Example: 1510
Given a parabola $(y  2)^2 = 8(x  1)$ What is the equation of its directrix?
A. x = 3
B. x = 1
C. y = 3
D. y = 3
\[(y  2)^2 = 8(x  1)\]
\[Y^2 = 8X\]
Here, $Y = y2, X = x1, 4a = 8, a = 2$
The equation of the directrix will be X = a or X = 2.
\[x1 = 2 \ \mathrm{or,}\ \ x = 1\]
Correct Answer: B
Circle
Learning Objectives:

Find equation of circle in standard and nonstandard form.

Given the equation of a circle, find its center and radius or diameter.
Equation of circle with center at Origin:
Equation of a circle with center at (0,0) and radius = r \[x^{2} + y^{2} = r^{2}\]
Equation of circle with center at point other that Origin:
Equation of a circle with center at (h,k) and radius = r
\[(xh)^{2} + (yk)^{2} = r^{2}\]
or in another form: \[x^{2} + y^{2} + 2gx + 2fy + c = 0\] which has center at (g,f) and
\[\mathrm{radius} = \sqrt{g^{2} + f^{2}  c}\]
A given equation in x and y can represent a circle, if it follows following three conditions:

The highest power of x is 2. Also, the highest power of y term is 2.

Coefficient of \(x^2\) is same as coefficient of \(y^2\).

There is NO product term xy.
Solved Example: 1301
Consider an ant crawling along the curve $(x2)^2 + y^2 = 4$ where x and y are in meters. The ant starts at the point (4,0) and moves counterclockwise with a speed of 1.57 meters per second. The time taken by the ant to reach the point (2,2) is (in seconds):
A. 1
B. 2
C. 4
D. $\pi/2$
$(x2)^2 + y^2 = 4$ is a circle with (2,0) as center and radius = 2. Its extreme point on east side is (4,0) and northmost point is (2,2), which means it will have to cover quarter circumference = $\pi \dfrac{r}{2}$ = $\pi$ m.
Time required = $\dfrac{d}{t} = \dfrac{\pi}{1.57}$ = 2 sec.
Correct Answer: B
Solved Example: 1302
What is the radius of the circle $x^2 + y^2  6y = 0?$
A. 2
B. 3
C. 4
D. 5
$x^2 + y^2  6y = 0$,
After completing the squares we get,
\begin{align*} x^2 + y^2  6y + 9 &= 9\\ x^2 + (y  3)^2 &= 3^2\\ \end{align*} Center = (0,3), Radius = 3Correct Answer: B
Solved Example: 1303
What are the coordinates of the center of the curve $x^2 + y^2  2x  4y  31 = 0?$
A. (1, 1)
B. (2, 2)
C. (1, 2)
D. (2, 1)
First Method: \begin{align*} x^{2}+y^{2}2x4y31 &=0\\ x^{2}2x+y^{2}4y &=31\\ x^{2}2x+1+y^{2}4y+4 &=31+1+4\\ \left( x1\right) ^{2}+\left( y2\right) ^{2} &=6^{2}\\ \end{align*} \[C\equiv \left( 1,2\right),\ r=6\] Second Method: \[x^{2}+y^{2}+2gx+2fy+c=0\] \[g=1,f=2,c=31\] \[C\equiv \left( g,f\right) =\left( 1,2\right)\] \begin{align*} r &=\sqrt {g^{2}+f^{2}c}\\ &=\sqrt {\left( 1\right) ^{2}+\left( 2\right) ^{2}\left( 31\right) }\\ &=\sqrt {36} = 6 \end{align*}Correct Answer: C
Solved Example: 1304
A circle whose equation is: $x^2 + y^2 + 4x +6y  23 = 0$ has its center at:
A. (2, 3)
B. (3, 2)
C. (3, 2)
D. (2, 3)
\[x^{2}+y^{2}+4x+6y23=0\] \[g=2,f=3\] \[C\equiv \left( g,f\right) =\left( 2,3\right)\]Correct Answer: D
Solved Example: 1305
What is the radius of a circle with the equation: $x^2  6x + y^2  4y  12 = 0$
A. 3.46
B. 7
C. 5
D. 6
\[x^{2}+y^{2}6x4y12=0\] \[g=3,f=2,c=12\] \begin{align*} r &=\sqrt {g^{2}+f^{2}c} \\&=\sqrt {\left( 3\right) ^{2}+\left( 2\right) ^{2}+12} \\&=\sqrt {9+4+12} \\&=\sqrt {25} = 5 \end{align*}Correct Answer: C
Solved Example: 1306
The diameter of a circle described by $9x^2 + 9y^2 = 16$
A. $\dfrac{4}{3}$
B. $\dfrac{16}{9}$
C. $\dfrac{8}{3}$
D. 4
$9x^{2}+9y^{2}=16, x^{2}+y^{2}=\dfrac {16}{9}$
Center $=\left( 0,0\right)$,
Radius $= \sqrt{\dfrac {16}{9}} = \dfrac {4}{3}$
Diameter = $2 \times \mathrm{radius} =2 \times \dfrac {4}{3}=\dfrac {8}{3}$
Correct Answer: C
Solved Example: 1307
How far from the yaxis is the center of the curve $2x^2 + 2y^2 +10x  6y  55 = 0$
A. 2.5
B. 3.0
C. 2.75
D. 3.25
\begin{align*} 2x^{2}+2y^{2}+10x6y55 &=0\\ x^{2}+y^{2}+5x3y\dfrac {55}{2} &=0 \end{align*} \[g=\dfrac {5}{2},f=\dfrac {3}{2},c=\dfrac {55}{2}, C\equiv \left( g,f\right) =\left( \dfrac {5}{2},\dfrac {3}{2}\right)\] \[\mathrm{ydistance}=\left x\right =\dfrac{5}{2}\]Correct Answer: A
Solved Example: 1308
What is the distance between the centers of the circles $x^2 + y^2 + 2x + 4y  3 = 0$ and $x^2 + y^2  8x  6y + 7 = 0?$
A. 7.07
B. 7.77
C. 8.07
D. 7.87
First circle center = (1,2), second circle center = (4,3). Using distance formula, the distance can be calculated.
Correct Answer: A
Solved Example: 1309
The shortest distance from A (3, 8) to the circle $x^2 + y^2 + 4x  6y = 12$ is equal to?
A. 2.1
B. 2.3
C. 2.5
D. 2.7
$x^2 + y^2 + 4x  6y = 12$, Center = (2,3), radius =5
Distance from (3,8) to the center (2,3) using distance formula = $\sqrt{{(3(2)}^2 + {(83)}^2}$ = $\sqrt{50}$ = 7.07
For shortest distance subtract the radius = 7.07  5 =2.07
Correct Answer: A
Solved Example: 1310
The equation $x^2 + y^2  4x + 2y  20 = 0$ describes:
A. A circle centered at the origin.
B. An ellipse centered at (2, 1).
C. A sphere centered at the origin.
D. A circle centered at (2, 1).
We will use completing the square method, \begin{align*} x^{2}+y^{2}4x+2y20 &=0\\ x^{2}4x+4+y^{2}+2y+1 &=20+5\\ \left( x2\right) ^{2}+\left( y+1\right) ^{2} &=25 \end{align*} This is a circle with, center $=\left( 2,1\right)$ and Radius = 5
Correct Answer: D
Solved Example: 1311
Find the area (in square units) of the circle whose equation is $x^2 + y^2 = 6x  8y$
A. 20 $\pi$
B. 22 $\pi$
C. 25 $\pi$
D. 27 $\pi$
\begin{align*} x^{2}+y^{2}&=6x8y\\ x^{2}+y^{2}6x+8y &=0 \end{align*} Comparing with the equation, \[x^{2}+y^{2}+2gx+2fy+c=0\] \[2g=6,g=3\] \[2f=8,f=4\] \[c=0\] \[r =\sqrt {g^{2}+f^{2}c} =\sqrt {\left( 3\right) ^{2}+4^{2}}=5\] \[\mathrm{Area} =\pi r^{2}=25\pi\ \mathrm{sq.\ units}\]Correct Answer: C
Solved Example: 1312
Determine the equation of the circle whose radius is 5, center on the line x = 2 and tangent to the line 3x  4y + 11 = 0.
A. $(x  2)^2$ +$ (y  2)^2$ = 5
B. $(x  2)^2$ + $(y + 2)^2$ = 25
C. $(x  2)^2$ + $(y + 2)^2 $= 5
D. $(x  2)^2$ +$ (y  2)^2$ = 25
\[r=5\] \[\left( x,y\right) =\left( 2,y\right)\] \[\left \dfrac {3\left( 2\right) 4\left( y\right) +11}{5}\right =5\] \[174y=25\] \[4y=8\] \[y=2\] So, the equation of the circle will be, $\left( x2\right) ^{2}+\left( y+2\right) ^{2}=25$Correct Answer: B
Solved Example: 1313
Find the equation of the circle with the center at (4, 5) and tangent to the line 2x + 7y  10 = 0.
A. $x^2$ + $y^2$ + 8x  10y  12 = 0
B. $x^2$ + $y^2$ + 8x  10y + 12 = 0
C. $x^2$ + $y^2$ + 8x + 10y  12 = 0
D. $x^2$ + $y^2$ – 8x + 10y + 12 = 0
\begin{align*} \left \dfrac {2\left( 4\right) +7\left( 5\right) 10}{\sqrt {2^{2}+7^{2}}}\right &=r\\ \left \dfrac {83510}{\sqrt {53}}\right &=r\\ \left \sqrt {53}\right &=r\\ r^{2} &=53 \end{align*} \begin{align*} \left( x+4\right) ^{2}+\left( y+5\right) ^{2} &=53\\ x^{2}+8x+16+y^{2}+10y+4 &=53\\ x^{2}+y^{2}+8x+10y12 &=0 \end{align*}
Correct Answer: C
Solved Example: 1314
Find the value of k for which the equation $x^2 + y^2 + 4x  2y  k = 0$ represents a point.
A. 5
B. 6
C. 6
D. 5
A point can be considered as circle with zero radius. \begin{align*} x^2 + y^2 + 4x  2y  k &= 0\\ x^{2}+4x+4+y^{2}2y+1k &=5\\ \left( x+2\right) ^{2}+\left( y1\right) ^{2}&=k+5 \end{align*} For point circle, r=0, $k+5 =0,\ \mathrm{Or,\ } k =5$
Correct Answer: D
Solved Example: 1315
The equation of a sphere with center at (3, 2, 4) and of radius 6 units is?
A. $x^2$ + $y^2$ +$ z^2 $+6x  4y  8z = 36
B. $ x^2$ +$ y^2 $+ $z^2$ +6x  4y  8z = 7
C. $x^2$ +$ y^2$ + $z^2$ +6x  4y + 8z = 6
D. $x^2$ + $y^2$ + $z^2$ +6x  4y + 8z = 36
\begin{align*} \left[ x\left( 3\right) \right] ^{2}+\left[ y2\right] ^{2}+\left[ z4\right] ^{2} &=6^{2}\\ \left( x+3\right) ^{2}+\left( y2\right) ^{2}+\left( z4\right) ^{2} &=36\\ x^{2}+6x+9+y^{2}4y+4+z^{2}8z+16 &=36\\ x^{2}+y^{2}+z^{2}+6x4y8z &=7 \end{align*}
Correct Answer: B
Ellipse
Learning Objectives:

Find equation of ellipse – types, regular and rectangular, eccentricity, directrix, focus, latus rectum, relation between a and b.
Equation of an ellipse when center is at (h,k) is:
\[\dfrac{(xh)^{2}}{a^{2}}+\dfrac{(yk)^{2}}{b^{2}} = 1\]
Equation of an ellipse when center is at (0,0) is: \[\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}} = 1\]
Foci: (\(\pm\)ae,0) = (\(\pm\) c, 0) where e is the eccentricity of the ellipse. \[c^2 = a^2  b^2\]
Here, a is called the semi major axis, and b is called semi minor axis. (Assuming a \(>\) b) For the ellipse shown in the figure, the semi major axis= 3 and semi minor axis = 1.
\[e = \sqrt{1  \dfrac{b^2}{a^2}} = \dfrac{c}{a}\]
Also, for an ellipse, e < 1
Solved Example: 1401
The general equation of a conic section is given by the following equation: $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ A curve maybe identified as an ellipse by which of the following conditions?
A. $B^2 $ 4AC < 0
B. $B^2$  4AC = 0
C. $B^2$  4AC > 0
D. $B^2$  4AC = 1
If B\(^2\)  4AC < 0 then the curve is an ellipse or circle.
If B\(^2\)  4AC = 0 then the curve is a parabola.
If B\(^2\)  4AC > 0 then the curve is a hyperbola.
Correct Answer: A
Solved Example: 1402
Point P (x, y) moves with a distance from point (0, 1) onehalf of its distance from line y = 4. The equation of its locus is?
A. 2$x^2$  4$y^2$ = 5
B. 4$x^2$ + 3$y^2$ = 12
C. 2$x^2$ + 5$y^3$ = 3
D. $x^2$ + 2$y^2$ = 4
\[\sqrt {\left( x0\right) ^{2}+\left( y1\right) ^{2}}=\dfrac {1}{2}\left( 4y\right)\] \[x^{2}+y^{2}2y+1=\dfrac {1}{4}\left( 4y\right) ^{2}\] \[4x^{2}+4y^{2}8y+4=168y+y^{2}\] \[4x^{2}+3y^{2}=12\]
Correct Answer: B
Solved Example: 1403
What is the length of the latus rectum of $4x^2 + 9y^2 + 8x  32 = 0?$
A. 2.5
B. 2.7
C. 2.3
D. 2.9
\begin{align*} 4x^{2}+9y^{2}+8x32&=0\\ 4\left( x^{2}+2x+1\right) +9y^{2}324&=0\\ \dfrac {\left( x+1\right) ^{2}}{9}+\dfrac {y^{2}}{4}&=1 \end{align*} \[a=3,b=2, \mathrm{Latus\ Rectum}=\dfrac {2b^{2}}{a} =\dfrac {2\left( 4\right) }{3}=2.66\]Correct Answer: B
Solved Example: 1404
The lengths of the major and minor axes of an ellipse are 10 m and 8 m, respectively. Find the distance between the foci.
A. 3
B. 4
C. 5
D. 6
Major axis = 2a =10, a=5, Minor axis =2b=8, b=4For ellipse, \begin{align*} b^{2}&=a^{2}\left( 1e^{2}\right)\\ 16&=25\left( 1e^{2}\right)\\ \dfrac {16}{25}&=1e^{2}\\ e^{2}&=1\dfrac {16}{25}=\dfrac {2}{25}, e=\dfrac {3}{5} \end{align*} Distance between foci, $=2ae =2\left( 5\right) \left( \dfrac {3}{5}\right) = 6$
Correct Answer: D
Solved Example: 1405
The equation $25x^2 + 16y^2  150x + 128y + 81 = 0$ has its center at?
A. (3, 4)
B. (3, 4)
C. (4, 3)
D. (3, 5)
\[25x^{2}+16y^{2}150x+128y+81=0\] To complete the squares we need the terms 225 and 256 respectively, these terms are added on both sides, \begin{align*} \left(25x^{2}150x+225\right)\\ +\left(16y^{2}+128y+256\right) &=81+225+256\\ \left( 5x15\right) ^{2}+\left( 4y+16\right) ^{2}&=400\\ \left( 5x15\right) ^{2}+\left( 4y+16\right) ^{2}&=20^{2} \end{align*} Center $=\left( \dfrac {15}{5},\dfrac {16}{4}\right) =\left( 3,4\right) $
Correct Answer: A
Solved Example: 1406
Find the major axis of the ellipse $x^2 + 4y^2  2x  8y + 1 = 0$
A. 2
B. 10
C. 4
D. 6
\begin{align*} x^2 + 4y^2  2x  8y + 1 &= 0\\ x^2 2x + 1 + 4y^2 8y &=0\\ (x1)^2 + 4(y^2  2y +1) &= 4\\ \dfrac{(x1)^2}{1} + \dfrac{(y1)^2}{\dfrac{1}{4}} &= 4\\ \dfrac{(x1)^2}{4} + \dfrac{(y1)^2}{1} &= 1 \end{align*} Comparing with standard equation, \[a^2 =4, or,\ a =2\] Major axis = 2a = 4Correct Answer: C
Solved Example: 1407
An ellipse with an eccentricity of 0.65 and has one of its foci 2 units from the center. The length of the latus rectum is nearest to?
A. 3.5
B. 3.8
C. 4.2
D. 3.2
\[e=0.65\] \[\mathrm{Focus}=\left( ae,0\right) =\left( 2,0\right)\] \[ae=2\] \[a=\dfrac {2}{e}=\dfrac {2}{0.65}=3.077\] \[b^{2}=a^{2}\left( 1e^{2}\right)\] \[b^{2}=5.467, b=2.338\] Length of latus rectum = $\dfrac {2b^{2}}{a}=\dfrac {2\left( 5.467\right) }{3.077}=3.55$
Correct Answer: A
Solved Example: 1408
The earth's orbit is an ellipse with the sun at one of the foci. If the farthest distance of the sun from the earth is 105.5 million km and the nearest distance of the sun from the earth is 78.25 million km, find the eccentricity of the ellipse.
A. 0.15
B. 0.25
C. 0.35
D. 0.45
\[2a=78.25+105.5\] \[a=91.875\] \[aae=78.25\] \[a\left( 1e\right) =78.25\] \[e=0.1482\simeq 0.15\]
Correct Answer: A
Solved Example: 1409
An ellipse with center at the origin has a length of major axis 20 units. If the distance from center of ellipse to its focus is 5, what is the equation of its directrix?
A. x = 18
B. x = 20
C. x = 15
D. x = 16
Given: Major axis = 2a = 20, a = 10, Also, focus = (ae, 0 ) = (5, 0) \begin{align*} a e &= 5\\ 10 e &= 5\\ e &= \dfrac{1}{2} \end{align*} Equation of directrix is given by, \[x = \dfrac{a}{e}= \dfrac{10}{\left(\dfrac{1}{2}\right)} = 20\]
Correct Answer: B
Hyperbola
Learning Objectives:

Find equation of hyperbola regular and rectangular, eccentricity, directix, focus, latus rectum, asymptote, relation between a and b.
Equation of a hyperbola when center is at (h,k) \[\dfrac{(xh)^{2}}{a^{2}} \dfrac{(yk)^{2}}{b^{2}} = 1\] Equation of a hyperbola when center is at (0,0) \[\dfrac{x^{2}}{a^{2}} \dfrac{y^{2}}{b^{2}} = 1\]

Focus of hyperbola: The two points on the transverse axis. These points are what controls the entire shape of the hyperbola since the hyperbola’s graph is made up of all points, P, such that the difference of distances between P and the two foci are equal. To determine the foci you can use the formula: \(a^2 + b^2 = c^2\)

Transverse axis: This is the axis on which the two foci are.

Asymptotes: The two lines that the hyperbolas come closer and closer to touching. The equation of the asymptotes is always: \[y = \pm \frac{a}{b} x\]
Solved Example: 1601
Find the equation of the hyperbola having vertices: ($\pm$5,0), foci: ($\pm$8,0)
A. $\dfrac{x^2}{39}  \dfrac{y^2}{25} =1$
B. $\dfrac{x^2}{14}  \dfrac{y^2}{25} =1$
C. $\dfrac{x^2}{25} + \dfrac{y^2}{39} =1$
D. $\dfrac{x^2}{25}  \dfrac{y^2}{39} =1$
For hyperbola,
$\mathrm{vertices} = (\pm a,0) = (\pm5,0)$,
$\mathrm{foci} = (\pm c,0), c^2 = a^2 + b^2$,
$\mathrm{\ or,\ }b^2 = 8^2  5^2 = 64  25 = 39$
Equation of a hyperbola, $\dfrac{x^2}{a^2}  \dfrac{y^2}{b^2} = 1 , \dfrac{x^2}{25}  \dfrac{y^2}{39} =1$
Correct Answer: D
Solved Example: 1602
What is the equation of the asymptote of the hyperbola $\dfrac{x^2}{9}  \dfrac{y^2}{4} =1$?
A. 2x  3y = 0
B. 3x  2y = 0
C. 2x  y = 0
D. 2x + y = 0
$a = 3, b = 2$. Equation of the asymptote is: $y = \pm \dfrac{b}{a}x, y = \pm \dfrac{2}{3}x, 3y = 2x, 2x  3y = 0$
Correct Answer: A
Solved Example: 1603
Find the equation of the hyperbola whose asymptotes are y = $\pm$ 2x and which passes through (5/2, 3).
A. 4$x^2$ + $y^2$ + 16 = 0
B. 4$x^2$ + $y^2$  16 = 0
C. $x^2$  4$y^2$  16 = 0
D. 4$x^2$  $y^2$ = 16
The equation of the asymptote is, $y = \dfrac{b}{a}x$.
Comparing with the given equation,$\dfrac{b}{a} = 2, b=2a$
So the equation of the hyperbola will become, $\dfrac{x^{2}}{a^{2}}\dfrac{y^{2}}{(2a)^{2}} = 1$
Substituting the given coordinates,
\[\dfrac{(5/2)^{2}}{a^{2}}\dfrac{3^{2}}{(2a)^{2}} = 1, a = 2, b= 4\]
\[\dfrac{x^{2}}{2^{2}}\dfrac{y^{2}}{4^{2}} = 1, 4x^2  y^2 = 16\]
Correct Answer: D
Solved Example: 1604
4$x^2$  $y^2$ = 16 is the equation of a/an?
A. Parabola
B. Hyperbola
C. Circle
D. Ellipse
\[4x^{2}y^{2}=16, \dfrac {x^{2}}{4}\dfrac {y^{2}}{16}=1\] Because of the negative sign of $y^2$ term this is an equation of a hyperbola in the format, $\dfrac {x^{2}}{a^{2}}\dfrac {y^{2}}{b^{2}}=1$
Correct Answer: B
Solved Example: 1605
Find the eccentricity of the curve: $9x^2  4y^2  36x + 8y = 4$
A. 1.80
B. 1.92
C. 1.86
D. 1.76
\begin{align*} 9x^{2}36x4y^{2}+8y &=4\\ 9\left( x^{2}4x+4\right) 364\left( y^{2}2y+1\right) +4 &=4\\ \dfrac {9}{36}\left( x2\right) ^{2}\dfrac {4}{36}\left( y1\right) ^{2} &=\dfrac {36}{36}\\ \dfrac {\left( x2\right) ^{2}}{4}\dfrac {\left( y1\right) ^{2}}{9} &=1 \end{align*} a = 2, b =3 \[b^{2} =a^{2}\left( e^{2}1\right), 9 =4\left( e^{2}1\right), 2.25 =e^{2}1, e =1.80\]
Correct Answer: A
Solved Example: 1606
How far from the xaxis is the focus F of the hyperbola $x^2  2y^2 + 4x + 4y + 4 = 0?$
A. 4.5
B. 3.4
C. 2.7
D. 2.1
\begin{align*} x^{2}2y^{2}+4x+4y+4 &=0\\ x^{2}+4x+42y^{2}+4y&=0\\ \left( x+2\right) ^{2}2\left( y^{2}2y\right) &=0\\ \left( x+2\right) ^{2}2\left( y^{2}2y+1\right) &=2\\ \left( x+2\right) ^{2}2\left( y1\right) ^{2}&=2\\ 2\left( y1\right) ^{2}\left( x+2\right) ^{2}&=2\\ \left( y1\right) ^{2}\dfrac {\left( x+2\right) ^{2}}{2}&=1\\ \end{align*} Comparing with, $\dfrac {Y^{2}}{a^{2}}\dfrac {X^{2}}{b^{2}}=1, a=1, \ b=\sqrt {2}$ \[X = x+2\ ,\ Y = y1\] \[e =\dfrac {\sqrt {a^{2}+b^{2}}}{a} =\dfrac {\sqrt {1+2}}{1} =\sqrt {3},\] \[F\equiv \left( 0,\pm ae\right) =\left( 0,\pm \sqrt {3}\right)\] \[X = x+2 \mathrm{\ Focus\ Shifted\ by}\ 2 \mathrm{\ in\ } x \mathrm{\ axis},\] \[Y = y1 \mathrm{\ Focus\ Shifted\ by}\ 1 \mathrm{\ in\ } y \mathrm{\ axis}\] \[\mathrm{Focus\ Shifted\ by}=\left( 2,1\right)\] \[F_{1} \equiv \left( 2,1+\sqrt {3}\right) =\left( 2,2.73\right),\] \[F_{2}\equiv \left( 2,1\sqrt {3}\right) =\left( 2,0.73\right)\]
Correct Answer: C
Distance Formula
Learning Objectives:

Calculate distance between two points in space.
Distance between two points in the space \(P_1\) \((x_1,y_1,z_1)\) and \(P_2\) \((x_2,y_2,z_2)\) is given by:
\[\sqrt{(x_{2} x_{1})^{2} + (y_{2} y_{1})^{2} + (z_{2} z_{1})^{2}}\]
This can be proved by repeated application of the Pythagorean Theorem.
Solved Example: 1701
The distance between the points (5,5,7) and (3,0,5) is:
A. 4.531
B. 5.657
C. 9.643
D. 93
\[d =\sqrt {\left( 3(5)\right) ^{2}+\left( 0(5)\right) ^{2}+\left( 57\right) ^{2}} =\sqrt {93} =9.643\]Correct Answer: C
Solved Example: 1702
If the distance between the points (3, 4) and (a, 2) is 8 units then find the value of a: (Airforce Group X Nov 2020)
A. $3 \pm 2\sqrt {15}$
B. $2\pm 2\sqrt {15}$
C. $1 \pm\sqrt {15}$
D. None of these
Correct Answer: A
Logarithms
Learning Objectives:
Convert between exponential and logarithmic form.
Use properties of logarithms.
Expand and condense logarithms.
Logarithm can be seen as opposite of exponent.
lf \(a^{b}=x\) then, \(log_{a}x=b\)
Multiplication property \[\log \left(xy \right) =\log x+\log y\]
Division property \[\log \left( \dfrac{x}{y}\right) =\log x  \log y\]
Power Property \[\log x^{y}=y\log x\]
Change of base property \[\log _{a}b=\dfrac{\log _{c}b}{\log _{c}a}\]
Solved Example: LOG01
Which of the following statements is not correct?
A. $\log_{10} 10 = 1$
B. $\log (2 + 3) = \log (2 \times 3)$
C. $\log_{10} 1 = 0$
D. $\log (1 + 2 + 3) = \log 1 + \log 2 + \log 3$
There is no such identity represented in (B). The correct form of
that expression will be: \[\log (2 * 3) =
\log 2 + \log 3\] (D) is not technically correct for any other
numbers. It is just true because 1 + 2 + 3 happens to be = 1 \(\times\) 2 \(\times\) 3, so eventually it becomes: \[\log (1 + 2 + 3) = \log 6 = \log (1 \times 2
\times 3) = \log 1 + \log 2 + \log 3\] Though it is a true
statement, it cannot be generalized as an identity.
Correct Answer: B
Solved Example: LOG02
If $\log 2$ = 0.3010 and $\log 3$ = 0.4771, the value of $\log_5 512$ is
A. 2.870
B. 2.967
C. 3.876
D. 3.912
\[\begin{aligned} \log_{5}512 &=\dfrac{\log 512}{\log 5}\\ &=\dfrac{\log 2^{9}}{\log 5}\\ &=\dfrac{9\log 2}{\log \left( \dfrac{10}{2}\right) }\\ &=\dfrac{9\log 2}{\log10\log 2}\\ &=\dfrac{9\log 2}{1\log 2}\\ &=\dfrac{9\left( 0.3010\right) }{1\left( 0.3010\right) }\\ &=3.876\end{aligned}\]
Correct Answer: C