Match the column I with column II and mark the correct option from the codes given below. Column I Column II i ii iii iv v $$xy+a^2-a(x+y)$$ $$2x^2-72xy+23y^2-4x-28y-48=0$$ $$6x^2-5xy-6y^2+14x+5y+4=0$$ $$14x^2-4xy+11y^2-44x-58y+71=0$$ $$4x^2-4xy+y^2-12x+6y+9=0$$ p q r s Ellipse Hyperbola Parabola Pair of straight lines

(i)$$=$$s,(ii)$$=$$q,(iii)$$=$$s,(iv)$$=$$p,(v)$$=$$s

(i)$$=$$p,(ii)$$=$$r,(iii)$$=$$q,(iv)$$=$$r,(v)$$=$$s

(i)$$=$$q,(ii)$$=$$p,(iii)$$=$$r,(iv)$$=$$s,(v)$$=$$q

(i)$$=$$r,(ii)$$=$$s,(iii)$$=$$p,(iv)$$=$$q,(v)$$=$$r

MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 11

From a point P perpendiculars PM and PN are drawn to curve

$3x^2+4xy+y^2=0$ to meet at M and N. If MN

$=2\sqrt{5}$ , then?

0%
Locus of P is pair of lines
0%
Locus of P is circle
0%
Area bounded by locus of P and x-axis above the x-axis is $50\pi$
0%
Distance of any tangent to locus of P from origin is $10$ unit

A circle has ccentre

$C$ on axes of parabola and it touches the parabola at point

$P$ .

$CP$ makes an angle of

$120^{o}$ with axis of parabola. If radius of circle is

$2$ , then latus rectum of parabola is

0%
$2$
0%
$4$
0%
$8$
0%
$16$

${ e }_{ 1 }$ is the eccentricity of the ellipse

$\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1and{ \quad e }_{ 2 }$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $${ e }_{ 1 }{ e }_{ 2 }=1$, then equation of the hyperbola is ___________________.

0%
$\dfrac { { x }^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 16 } =1$
0%
$\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$
0%
$\dfrac { { x }^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 25 } =1$
0%
none of these

Length of latus rectum of the parabola

$9x^{2}+16y^{2}+24xy-4x+3y=0$ is :

0%
$\dfrac{1}{20}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{5}$
0%
$1$

Find the equation of the circle having

$(1, -2)$ as its centre and passing through the intersection of the lines

$3x+y=14$ and

$2x+5y=18$ .

0%
$x^2-y^2+2x+4y+20=0$
0%
$x^2+y^2+2x-4y+20=0$
0%
$x^2+y^2-2x+4y-20=0$
0%
$x^2-y^2+2x-4y+20=0$

The equation of the circle which bisects the circumference of the circles

$x^{2}+y^{2} =1, \;x^{2}+y^{2}-2x =3$ and

$x^2+y^{2}+2y =3$ is

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

The equation of the circle passing through

$(4,6)$ and having centre

$(1,2)$ is

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

Explanation

Let coordinates of point A are

$\left( 4,6 \right)$

As circle is passing through point A, point A lies on circle.

Let center of circle is

$C\left( 1,2 \right)$

$\therefore h=1$ and

$k=2$

Thus, AC is radius of circle.

By distance formula,

$AC=r=\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 6-2 \right) }^{ 2 } }$

$\therefore r=\sqrt { { \left( 3 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } }$

$\therefore r=\sqrt { 9+16 }$

$\therefore r=\sqrt { 25 }$

$\therefore r=5$

Thus, equation of circle is,

${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ r }^{ 2 }$

${ \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }=\left( 5 \right) ^{ 2 }$

${ x }^{ 2 }-2x+1+y^{ 2 }-4y+4=25$

$\therefore { x }^{ 2 }+y^{ 2 }-2x-4y-20=0$

Thus, answer is option (A)

If the length of the latus rectum of the parabola

$289(x-3)^{2}+289(y-1)^{2}=(15x-8y+13)^{2}$ is

$k$ , then

$[k]=?$

0%
$5$
0%
$2$
0%
$3$
0%
$1$

The latus rectum of a conic section is the width of the function through the focus. The positive difference between the length of the latus rectum of

$3y =x^{2}+4x-9$ and

$x^{2}+4y^{2}-6x+16y =24$ is-

0%
$\frac{1}{2}$
0%
2
0%
$\frac{3}{2}$
0%
$\frac{5}{2}$

If the line

$3x+4y=1$ touches the parabola

$y^{2}=4ax$ then length of its latus rectum is equal to

0%
$3/16$
0%
$3/8$
0%
$3/4$
0%
$9/16$

The parabola

$y^{6}=4ax$ passes through the point

$(2,-6)$ , then the length of its latus rectum is

0%
$18$
0%
$9$
0%
$6$
0%
$16$

The length of the latus rectum of the parabola

$x^2=ay+by+c$ is

0%
$\dfrac{a}{4}$
0%
$\dfrac{a}{3}$
0%
$\dfrac{1}{a}$
0%
$\dfrac{1}{4a}$

A tangent drawn to hyperbola

$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ at

$P\left(\dfrac{\pi}{6}\right)$ forms a triangle of area

$3a^{2}$ square units, with coordinate axes. Its eccentricity is equal to

0%
$4$
0%
$5$
0%
$\dfrac{9}{2}$
0%
$\sqrt{17}$

In the line

$2x+3y=1$ touches the parabola

$y^{2}=4a(x+a)$ , then the length of its latus rectum is

0%
$\dfrac {8}{9}$
0%
$\dfrac {8}{13}$
0%
$4$
0%
$\dfrac {4}{9}$
0%
$\dfrac {4}{13}$

If the latus rectum subtends a right angel at the center of a hyperbola, then its eccentricity is

0%
$\sqrt{2}$
0%
$\dfrac{\sqrt{3}+1}{2}$
0%
$\dfrac{\sqrt{5}-1}{2}$
0%
$\dfrac{\sqrt{3}+\sqrt{5}}{2}$

The equation

$y = \sqrt{2 - x^2 - y^2} + \sqrt{x^2 + y^2 - 2}$ represents, where

$x > 0$

0%
Circle with radius $2$ units
0%
Parabola
0%
A point circle
0%
Ellipse

If PQ is a double ordinate of the hyperbola

$\frac { { x }^{ 2 } }{ { a }^{ 2 } } \frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ , such that OPQ is an equilateral tnangle, O being the centre of the hyperbola, then the eccentncity e of the hyperbola satisfies

0%
$e=\frac { 2 }{ \sqrt { 3 } }$
0%
$e=\frac { \sqrt { 3 } }{ 2 }$
0%
$e>\frac { 2 }{ \sqrt { 3 } }$
0%
1

A tangent to the curve y = f(x) cuts the line y = x at a point which is at a distance of 1 unit from y - axis. The equation of the curve is

0%
$\dfrac{x - 1}{y - 1} = C$
0%
$\dfrac{x}{y} = C$
0%
$xy = C$
0%
none of these

The length of the latus rectum of the parabola

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

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

Eccentricity of the hyperbola whose asymptotes are given by

$3x + 2y +5 = 0$ and

$2x +3y+ 5 = 0$ is

0%
$\sqrt 2$
0%
$\dfrac{3}{2}$
0%
$2$
0%
None of these

Explanation

$\begin{array}{l} 3x+2y+5=0 & 2x+3y+5=0 \\ 3x+2y=-5 & -3y=2x-5 \\ 2y=-3x-5 & y=\dfrac { { -2 } }{ { \left( { -3 } \right) } } x-\dfrac { 5 }{ { \left( { -3 } \right) } } \\ y=\dfrac { { -3 } }{ 2 } x-\dfrac { 5 }{ 2 } & y=\dfrac { 2 }{ 3 } x+\dfrac { 5 }{ 3 } \\ {m }=\dfrac { { -3 } }{ 2 } \times \dfrac { 2 }{ 3 } =-1 & \end{array}$

It is a rectangle

$\therefore e=\sqrt { 2 }$

$x=2+3\cos\theta$ and

$y=1-3\sin\theta$ represent a circle then the centre and radius is?

0%
$(2, 1), 9$
0%
$(2, 1), 3$
0%
$(1, 2), \dfrac{1}{3}$
0%
$(-2, -1), 3$

Explanation

We have,

$x=2+3\cos\theta\Rightarrow 3\cos \theta\Rightarrow 3\cos\theta =x-2$ ..(i)

$y=1-3\sin\theta \Rightarrow 3\sin\theta =-y+1$ ..(ii)
Squaring adding (i) & (ii), we get

$(x-2)^2(-(y-1))^2=3^2(\cos^2\theta +\sin^2\theta)$

$\Rightarrow (x-2)^2+(-(y-1))^2=3^2$

$\therefore$ Radius

$=3$ , centre

$=(2, 1)$ .

The name of the conic whose focus is

$\left(-1,1\right)$ corresponding direction is

$x-y+1=0$ whose length of semi latusrectum is

$3$ , is

0%
Parabola
0%
Ellipse
0%
Hyperbola
0%
Pair of lines

The graph of the conic

$x ^ { 2 } - ( y - 1 ) ^ { 2 } = 1$ has one tangent line with positive slope that passes through the origin, the point of tangency being (a, b). Then Length of the latus rectum of the conic is

0%
$1$
0%
$\sqrt { 2 }$
0%
$2$
0%
None

Let PQ be a variable focal chord of the parabola

${ y }^{ 2 }=4ax$ where vertex is A. Locus of, centre triangle APQ is a parabola

$'{ P }_{ 1 }'$ Latus rectum of parabola

${ P }_{ 1 }$ is

0%
$\dfrac { 2a }{ 3 }$
0%
$\dfrac { 4a }{ 3 }$
0%
$\dfrac { 8a }{ 3 }$
0%
$\dfrac { 16a }{ 3 }$

If S be the focus of a parabola and PQ be the focal chord,such that SP=3 and SQ = 6,then the length of latus rectum of the parabola is

0%
4
0%
2
0%
8
0%
16

The equation of the image of the circle

${ \left( x-3 \right) }^{ 2 }\quad +\quad { \left( y-2 \right) }^{ 2 }\quad =1\quad in\quad the\quad mirror\quad x+y=19\quad is$

0%
${ \left( x-14 \right) }^{ 2 }\quad +\quad { \left( y-13 \right) }^{ 2 }\quad =1$
0%
${ \left( x-15 \right) }^{ 2 }\quad +\quad { \left( y-14 \right) }^{ 2 }\quad =1$
0%
${ \left( x-16 \right) }^{ 2 }\quad +\quad { \left( y-15 \right) }^{ 2 }\quad =1$
0%
${ \left( x-17 \right) }^{ 2 }+{ \left( y-16 \right) }^{ 2 }\quad =1$

If eccentricity of the hyperbola

$\dfrac {x^{2}}{\cos^{2}\theta}-\dfrac {y^{2}}{\sin^{2}\theta}=1$ is more then

$2$ when

$\theta\ \in \ \left(0,\dfrac {\pi}{2}\right)$ . Find the possible values of length of latus rectum

0%
$(3,\infty)$
0%
$(1,3/2)$
0%
$(2,3)$
0%
$(-3,-2)$

Let

$0<\theta <\dfrac { \pi }{ 2 } .$ If the eccentricity of the hyperbola

$\dfrac { { x }^{ 2 } }{ { cos }^{ 2 }\theta } -\dfrac { { y }^{ 2 } }{ { sin }^{ 2 }\theta } =1$ is greater than 2, then the length of its latus rectum lies in the interval:

0%
$(1,3/2]$
0%
$(3/2,2]$
0%
$\left( 3,\infty \right)$
0%
$(2,3]$

The lines

$2x-3y-5=0$ and

$3x-4y=7$ are diameters of a circle of area

$154 \ sq\ units$ , then the equation if the circle is :

0%
$x^2 +y^2 +2x-2y-62=0$
0%
$x^2 +y^2 +2x-2y-47=0$
0%
$x^2 +y^2 -2x+2y-62=0$
0%
$x^2 -y^2 -2x+2y-62=0$

Explanation

$\pi r^2=154$

$r=7$

For centre on slowing equation we get,

$2x-3y=5$ …………

$(1)$

$3x-4y=7$ ………….

$(2)$

Multiply

$3$ to equation

$(1)$ and Multiply

$2$ to equation

$(2)$

$6x-9y=15$

$6x-8y=14$

_____________

$-y=1$

$y=-1$

$x=1$

So centre is

$(1, -1)$

Equation of the circle

$(x-1)^2+(y+1)^2=7^2$

$\Rightarrow x^2+y^2-2x+2y=47$

$\Rightarrow x^2=y^2-2x+2y-47=0$ .

1221283_1315418_ans_ce563dae3d214276833aa06a13a3cef3.jpg

Axes are coordinates axes, the ellipse passes through the points where the straight line

$\dfrac {x}{4}+\dfrac {y}{3}=1$ meets the coordinates axes. Then equation of the ellipses is

0%
$\dfrac {x^{2}}{16}+\dfrac {y^{2}}{9}=1$
0%
$\dfrac {x^{2}}{64}+\dfrac {y^{2}}{36}=1$
0%
$\dfrac {x^{2}}{4}+\dfrac {y^{2}}{3}=1$
0%
$\dfrac {x^{2}}{8}+\dfrac {y^{2}}{6}=1$

The length of the latus rectum of the parabola,

$y^2-6y+5x=0$ is

0%
$1$
0%
$3$
0%
$5$
0%
$7$

The equation

${ x }^{ 2 }+{ y }^{ 2 }+4x+6y+13=0\quad$ represents

0%
Circle
0%
Pair of coincident straight lines
0%
Pair of concurrent straight lines
0%
Point

Length of the latus rectum of the parabola

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

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

Which of the following can be the equation of an ellipse?

0%
$x^{2} + y^{2} = 5$
0%
$\dfrac {x^{2}}{9} + \dfrac {x^{2}}{9} = 1$
0%
$2x^{2} + 3y^{2} = 5$
0%
$2x + 2y = 5$

The curve for which the normal at any point (x,y) and the line joining origin to that point form and isosceles triangle with the x-axis as base is

0%
an ellipse
0%
a rectangular hyperbola
0%
a circle
0%
circle.

The equation of circle at center

$(3,4)$ and radius

$5$ .

0%
$x^2+y^2-6x-8y=0$
0%
$x^2+y^2+3x+4y+5=0$
0%
$x^2+y^2+6x+8y=0$
0%
None.

Explanation

The center of equation

$(3,4)$

The radius of circle is

$5$

The equation of circle is

$(x-3)^2+(y-4)^2=5^2\\x^2+y^2-6x+9-8y+16=25\\x^2+y^2-6x-8y=0$

If two distinct chords of a parbola

${ y }^{ 2 }=4ax$ passing through (a , 2a) are bisected on the line x + y =1, then the length of the latnsrectum can be

0%
2
0%
1
0%
4
0%
3

If two distinct chords of a parabola

${ y }^{ 2 }=4ax$ passing through

$(a, 2a)$ are bisected on the

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

0%
2
0%
1
0%
4
0%
3

If the parabola

${y^2} = 4\,\,ax$ passes through the point

$\left( { - 3,2} \right),$ then the length of its latus rectum is

0%
2 / 3
0%
4 / 3
0%
1 / 3
0%
4

Match the column I with column II and mark the correct option from the codes given below.

	Column I		Column II
i ii iii iv v	$xy+a^2-a(x+y)$ $2x^2-72xy+23y^2-4x-28y-48=0$ $6x^2-5xy-6y^2+14x+5y+4=0$ $14x^2-4xy+11y^2-44x-58y+71=0$ $4x^2-4xy+y^2-12x+6y+9=0$	p q r s	Ellipse Hyperbola Parabola Pair of straight lines

0%
(i) $=$ s,(ii) $=$ q,(iii) $=$ s,(iv) $=$ p,(v) $=$ s
0%
(i) $=$ p,(ii) $=$ r,(iii) $=$ q,(iv) $=$ r,(v) $=$ s
0%
(i) $=$ q,(ii) $=$ p,(iii) $=$ r,(iv) $=$ s,(v) $=$ q
0%
(i) $=$ r,(ii) $=$ s,(iii) $=$ p,(iv) $=$ q,(v) $=$ r

$P S Q$ is the focal chord of the parabola

$y^{2}=8 x$ such that

$S P=6$ . Then the length of

$SQ$ is

0%
$6$
0%
$4$
0%
$3$
0%
none of these

$x ^ { 2 } + y ^ { 2 } - 2 x + 2 a y + a + 3 = 0$ represents a real circle with non-zero radius, then most appropriate is?

0%
$a \in ( - \infty , - 1 )$
0%
$a \epsilon ( - 1,2 )$
0%
$a \epsilon ( 2 , \infty )$
0%
$a \epsilon ( - \infty , - 1 ) \cup ( 2 , \infty )$

Equation of the circle whose radius is 3 and centre is

$(-1,2)$

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

If C is the centre and A, B are two points on the conic

${ 4x }^{ 2 }+{ 9y }^{ 2 }-8x-36y+4=0$ such that

$\angle ABC-\dfrac { \pi }{ 2 } ,$ then

$\begin{matrix} 1 & \quad \quad +\quad \quad \quad \quad \quad 1 \\ { CA }^{ 2 } & \quad \quad \quad \quad { CB }^{ 2 } \end{matrix}=$

0%
$\dfrac { 13 }{ 36 }$
0%
$\dfrac { 36 }{ 13 }$
0%
$\dfrac { 16 }{ 33 }$
0%
$\dfrac { 33 }{ 16 }$

If the equation

$p{ x }^{ 2 }+(2-q)xy+3{ y }^{ 2 }-6qx+30y+6q=0$ represents a circle, then the values of p and q are

0%
3, 1
0%
2, 2
0%
3, 2
0%
3, 4

The length of the lotus rectum of the parabola

$9x^{2}-6x+36y+19=0$

0%
$36$
0%
$9$
0%
$6$
0%
$4$

Equation of the circle whose radius is

$a+b$ and centre

$(a, -b)$

0%
${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=2ab$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=-2ab$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=ab$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=-ab$

The foci of a hyperbola with the foci of the ellipse

$\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 12 } =1.$ Find the equation of the hyperbola, if its eccentricity is 2.

0%
$\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1$
0%
$\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =2$
0%
$\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 12 } =1$
0%
None of these

$x^{2}+y^{2}-2 x+2 a y+a+3=0$ represents a real circle with non-zero radius, then most appropriate is

0%
$a \in(-\infty,-0)$
0%
$a \in(-1,2)$
0%
$a \in(2,\infty)$
0%
$a \in(-\infty,-1) \cup(2 \infty)$

Find the equation of a circle whose cemtre is

$\left( {2,1} \right)$ and radius is 3

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

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

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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