If $$y ^ { 2 } - 2 x - 2 y + 5 = 0$$ is

a circle with centre $$( 1,1 )$$

a parabola with directrix $$x = 3 / 2$$

MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 12

The latus rectum of parabola

${ y }^{ 2 }=5x+4y+1$ is

0%
10
0%
5
0%
$\dfrac { 5 }{ 4 }$
0%
$\dfrac { 5 }{ 2 }$

A circle touch the line L and the circle

${ C }_{ 1 }$ externally such that both the circles are on the same side of the line, then the locus of center of the circle is

0%
ellipse
0%
hyperbola
0%
parabola
0%
parts of straight line

The equations

$x=\dfrac{t}{4}, y=\dfrac{t^2}{4}$ represents

0%
An ellipse
0%
A parabola
0%
A circle
0%
A hyperbola

Length of the latus rectum of the hyperbola

$x y=c^{2},$ is

0%
$\sqrt{2} {c}$
0%
$2c$
0%
$2\sqrt{2}{c}$
0%
$4c$

Lissajous figure obtained by combining x=ASin

$\omega t$ and y=ASin

$(\omega t+\Pi /4)$ will be

0%
an ellipse
0%
a straight line
0%
a circle
0%
a parabola

Let

$PQ$ be a variable focal chord of the parabola

${y^2} = 4ax\left( {a > 0} \right)$ whose vertex is A. then the locus of centroid of

$\Delta APQ$ lies on a parabola whose length of latusrectum is

0%
$\frac{{2a}}{3}$
0%
a
0%
$\frac{{4a}}{3}$
0%
$\frac{{5a}}{3}$

Vertex of the parabola

$2{ y }^{ 2 }+3y+4x-1=0$ is

0%
$\left( \dfrac { 25 }{ 32 } ,\dfrac { -7 }{ 4 } \right)$
0%
$\left( \dfrac { 25 }{ 32 } ,\dfrac { -3 }{ 4 } \right)$
0%
$\left( \dfrac { 15 }{ 32 } ,\dfrac { 7 }{ 4 } \right)$
0%
$\left( \dfrac { 17 }{ 32 } ,\dfrac { -3 }{ 4 } \right)$

A circle passes through

$A(1,2)$ and the equations of the normal to the circle is

$x+2y=5$ . If the circle passes through

$B(-5,5)$ , then the radius of the circle is

0%
$\dfrac {3}{2}$
0%
$\dfrac {\sqrt {5}}{2}$
0%
$\dfrac {3\sqrt {5}}{2}$
0%
$\dfrac {5}{2}$

If the radius of the circle

${x}^{2}+{y}^{2}+2gx+2fy+c=0$ be

$r$ , then it will touch both the axes, if

0%
$g=f=c$
0%
$g=f=c=r$
0%
$g=f=\sqrt{c}=r$
0%
$g=f$ and ${c}^{2}=r$

If two distinct chords of a parabola

$y^2 = 4 \,ax$ passing through the point

$(a, 2a)$ are bisected by the line

$x + y = 1$ , then the length of the latus rectum can be

0%
$(2,6)$
0%
$(1,4)$
0%
$(0,2)$
0%
$(0,4)$

The equations

$x=\frac{t}{4},y=\frac{t^2}{4}$ represents

0%
An ellipse
0%
A parabola
0%
A circle
0%
A hyperbola

If the radius of the circle

$x ^ { 2 } + y ^ { 2 }$ - 18 x - 12 y + k = 0 be 11 then k =

0%
347
0%
4
0%
-4
0%
97

The latus rectum of a parabola whose focal chord is PSQ such that SP=3 and SQ=2 is given by

0%
24/5
0%
12/5
0%
6/5
0%
23/5

$\frac { (3x-4y-{ 1) }^{ 2 } }{ 100 } -\frac { (4x+3y-{ 1) }^{ 2 } }{ 225 } =1,$ then
length latusrectum of hyperbola is-

0%
$\frac { 9 }{ 2 }$
0%
$\frac { 40 }{ 3 }$
0%
9
0%
$\frac { 8 }{ 3 }$

$y ^ { 2 } - 2 x - 2 y + 5 = 0$ is

0%
a circle with centre $( 1,1 )$
0%
a parabola with vertex
0%
a parabola with directrix $x = 3 / 2$
0%
a parabola with directrix

The value of

$\alpha$ for which three distinct chords drawn from

$(\alpha ,0)$ to the ellipse

${ x }^{ 2 }+2{ y }^{ 2 }=1$ are bisected by the parabola

${ y }^{ 2 }=4x$ is

0%
$9$
0%
$\sqrt { 17 }$
0%
$8$
0%
None of these.

The equation

$\dfrac{x^2}{10 - a} + \dfrac{y^2}{4 - a} = 1$ , represents an ellipse , if

0%
$a < 4$
0%
$a > 4$
0%
$4 < a < 10$
0%
$a > 10$

Let PQ be the latus rectum of the parabola

$y^2=4x$ with vertex A. Minimum length of the projection of PQ on a tangent drawn in portion of parabola PAQ is :

0%
$2$
0%
$4$
0%
$2 \sqrt{3}$
0%
$2 \sqrt{2}$

$C:\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1$
The equation of parabolas with same latus- rectum as conic C, is/are

0%
${y^2} - 5x + 3 = 0$
0%
${y^2} +6x - 21 = 0$
0%
${y^2} - 6x - 21 = 0$
0%
${y^2} +6x + 3 = 0$

The distance between the foci or a hyperbola is double the distance between its vertices and the length of a conjugate axis isThe equation of the hyperbola referred to its axes as axes of coordinates is

0%
$3 x ^ { 2 } - y ^ { 2 } = 3$
0%
$x ^ { 2 } - 3 y ^ { 2 } = 3$
0%
$3 x ^ { 2 } - y ^ { 2 } = 9$
0%
$x ^ { 2 } - 3 y ^ { 2 } = 9$

The equation of directrix of a parabola is 3x + 4y + 15 =0 and equation of tangent at vertex is 3x + 4y - 5=Then the length of latus recturn is equal to

0%
15
0%
14
0%
13
0%
16

Length of the latus rectum of the parabola

$25\left[ { \left( x-2 \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 } \right] ={ \left( 3x-4y+7 \right) }^{ 2 }$ is

0%
4
0%
2
0%
1/5
0%
2/5

equation of latus rectum of the parabola

${ y }^{ 2 }-16x-6y+1=0$ is

0%
x-7=0
0%
x+7=0
0%
2x-7=0
0%
2x+7=0

The latus rectum of a parabola whose directrix is x + y- 2 =0 and focus is (3, -4) is

0%
$-3\sqrt { 2 }$
0%
$3\sqrt { 2 }$
0%
$2\sqrt { 2 }$
0%
$3\sqrt { 2 }$

The y-axis is the directrix of the ellipse with eccentricity e=1/2 and the corresponding focus is at (3, 0), equation to its auxilary circle is

0%
$x^2+y^2-8x+12=0$
0%
$x^2+y^2-8x-12=0$
0%
$x^2+y^2-8x+9=0$
0%
$x^2+y^2=4$

If e,e' be the eccentricities of two conics S and S' and if

$e^2+e^{,2}=3$ , then bothe S and S'

0%
Ellipses
0%
Parabola
0%
Hyperbola
0%
None of these

The equation of circle with centre (1, 2) and tangent

$x + y - 5 = 0$ is

0%
${x^2} + {y^2} + 2x - 4y + 6 = 0$
0%
${x^2} + {y^2} - 2x - 4y + 3 = 0$
0%
${x^2} + {y^2} - 2x + 4y + 8 = 0$
0%
${x^2} + {y^2} - 2x - 4y + 8 = 0$

The equation of the ellipse with axes along the x-axis and the y-axis, which passes through the points P(4, 3) and Q (6, 2) is

0%
$\frac{x^2}{50}+\frac{y^2}{13}=1$
0%
$\frac{x^2}{52}+\frac{y^2}{13}=1$
0%
$\frac{x^2}{13}+\frac{y^2}{52}=1$
0%
$\frac{x^2}{52}+\frac{y^2}{17}=1$

If the circle describe on the line joining the points (0, 1) and

$(\alpha ,\beta )$ as diameter cuts the x-axis in points whose abscissae are roots of equation

$x^2-x+3=0$ the

$(\alpha ,\beta )$

0%
(1, 3)
0%
(1, 5)
0%
(-5, 1)
0%
(-5, -1)

The equation of a circle with origin as a centre and passing through equilateral whose median is of length 3a is

0%
${ x }^{ 2 }+{ y }^{ 2 }={ 9a }^{ 2 }$
0%
${ x }^{ 2 }+{ y }^{ 2 }={ 16a }^{ 2 }$
0%
${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$
0%
None of these

If the parabola

${ y }^{ 2 }=4ax$ passes through (2,6) then the equation of the latusrectum is

0%
2x -9 = 0
0%
4x + 9 = 0
0%
2x + 9 = 0
0%
4x - 9 = 0

The length of the latus rectum of the parabola

$4 y ^ { 2 } + 2 x - 20 y + 17 = 0$ is

0%
$3$
0%
$6$
0%
$1 / 2$
0%
None

The length of the latus rectum of the parabola

${ y }^{ 2 }-4x+4y+8=0\quad is$

0%
8
0%
6
0%
4
0%
2

The equation

$\dfrac { { x }^{ 2 } }{ 12-a } +\dfrac { { y }^{ 2 } }{ 4-a } =1$ represent an ellipse, if:

0%
$a < 12$
0%
$a < 4$
0%
$a > 12$
0%
$a > 4$

The equation of circles passing through (3,-6) touching both the axes is

0%
$x^{2}+y^{2}-6x+6y+9=0$
0%
$x^{2}+y^{2}+6x-6y+9=0$
0%
$x^{2}+y^{2}-30x-30y+225=0$
0%
$x^{2}+y^{2}-30x+30y+225=0$

$a\neq b$ the parametric equations

$x=a(cos \Theta +sin \Theta ),y=b(cos\Theta -sin\Theta )$ represents

0%
Hyperbola
0%
A circle
0%
An ellipse
0%
A pair of straight lines
0%
A parabola

A common tangent to the conics

${ x }^{ 2 }=6y$ and

$2{ x }^{ 2 }-4{ y }^{ 2 }=9$ , is __________.

0%
$x+y=1$
0%
$x-y=1$
0%
$x+y=\dfrac { 9 }{ 2 }$
0%
$x-y=\dfrac { 3 }{ 2 }$

The latus rectum of a parabola whose focal chord is PSQ such that SP=3 and SQ=2 is given by

0%
24/5
0%
12/5
0%
6/5
0%
none of these

The eccentricity of the hyperbola

$x^{2}-3y^{2}+1=0$ is

0%
$\frac{1}{2}$
0%
1
0%
2
0%
3

$5x+9=0$ is the directrix of the hyperbola

$16{x}^{2}-9{y}^{2}=144$ , then its correponding focus is:

0%
$\left( -\cfrac { 5 }{ 3 } ,0 \right)$
0%
$(5,0)$
0%
$(-5,0)$
0%
$\left( \cfrac { 5 }{ 3 } ,0 \right)$

Explanation

$\textbf{Step 1 : Convert the given equation of hyperbola in parametric form}$

$Given : \,16x^2-9y^2=144$

$\implies \,\cfrac{16x^2}{144}-\cfrac{9y^2}{144}=1$

$\implies \,\cfrac{x^2}{9}-\cfrac{y^2}{16}=1\,\qquad\qquad...(i)$

$\implies a^2=9\,\implies a=3\quad ,b^2=16\qquad\qquad ...(ii)$

$\textbf{Step 2 : Write equation of directrix in standard form}$

$\textbf{Equation of directrix to the hyperbola }\,\boldsymbol{\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1 \,\textbf{is}}$

$\boldsymbol{x\pm\cfrac a e=0}$

$\text{Given equation of a directrix is}$

$5x+9=0$

$\text{dividing by 5 on both sides,}$

$\implies x+\cfrac 9 5=0$

$\because a=3,\qquad\qquad...[from\,eq^n \,(ii)]$

$\implies x+\cfrac {3} {\cfrac 5 3}=0$

$\therefore e=\cfrac 5 3 \,\text{ and directrix lies in left side of hyperbola.}$

$\text{Hence corresponding focus will be at negative }X-axis.$

$\therefore \textbf{focus is at } \boldsymbol{(-ae,0)}\equiv (-3)\times \cfrac 53\equiv (-5, 0)$

$\therefore \textbf{The focus corresponding to the directrix }\,\boldsymbol{5x+9=0\,\textbf{is at }\,(-5, 0) .}$

$\textbf{Hence, option C is correct .}$

If the centroid of an equilateral triangle is (1,1) and one of its vertices is (-1,2) then, equation of its circum circle is

0%
$x^{2}+y^{2}-2x-2y-3=0$
0%
$x^{2}+y^{2}+2x-2y-3=0$
0%
$x^{2}+y^{2}-4x-6y+9=0$
0%
$x^{2}+y^{2}+x-y+5=0$

The eccentricity of an ellipse, with its centre at the origin, is

$\frac{1}{2}$ . If one of the directrices is x = 4, then the equation of the ellipse is

0%
$3x^2 + 4y^2$ = 1
0%
$3x^2 + 4y^2$ = 12
0%
$4x^2 + 3y^2$ = 12
0%
$4x^2 + 3y^2$ = 1

Angle between the parabola

$y^{2} = 4b(x - 2a + b)$ and

$x^{2} + 4a(y - 2b - a) = 0$ at the common end of their latus rectum, is

0%
$\tan^{-1}(1)$
0%
$\cot^{-1}1 + \cot^{-1}\dfrac {1}{2} + \cot^{-1} \dfrac {1}{3}$
0%
$\tan^{-1} (\sqrt {3})$
0%
$\tan^{-1} (2) +\tan^{-1} (3)$

Equation of a circle whose centre is in

$I$ quadrant as

$\left(\alpha,\ \beta\right)$ and touches

$x-$ axis will be:

0%
$x^{2}+y^{2}-2\alpha x - 2\beta y + \alpha ^{2}$
0%
$x^{2}+y^{2}+2\alpha x - 2\beta y + \alpha ^{2}$
0%
$x^{2}+y^{2}-2\alpha x + 2\beta y + \alpha ^{2}$
0%
$x^{2}+y^{2} + 2\alpha x + 2\beta y + \alpha ^{2}$

Explanation

Given that: A circle centre in I quadrant as

$(\alpha,\beta)$ and touches x-axis.

To find: Equation of the circle.

Solution:

Refer image,

$\therefore$ Centre (c)

$=(\alpha,\beta)$

$\because\,CP$ is parallel to y-axis

$\therefore$ Radius (p)

$=CP=\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}$

$=\sqrt{(\alpha-\alpha)^2+(0-\beta)^2}$

$=\sqrt{\beta^2}$

$=\beta$

$\therefore$ Equation of the circle is

$(x-\alpha)^2+(y-\beta)^2=\beta^2$

$\Rightarrow x^2-2\alpha x+\alpha^2+y^2-2\beta y+\beta^2=\beta^2$

$\Rightarrow x^2+y^2-2\alpha x-2\beta y+\alpha^2=0$

Hence, the required is

$x^2+y^2-2\alpha x-2\beta y+\alpha^2=0$

1912310_1897877_ans_c91aa7601aeb478d912257504a13abdd.png

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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