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CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Conic Sections
Quiz 12
The latus rectum of parabola $${ y }^{ 2 }=5x+4y+1$$ is
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10
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5
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$$\dfrac { 5 }{ 4 } $$
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$$\dfrac { 5 }{ 2 } $$
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
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ellipse
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hyperbola
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parabola
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parts of straight line
The equations $$x=\dfrac{t}{4}, y=\dfrac{t^2}{4}$$ represents
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An ellipse
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A parabola
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A circle
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A hyperbola
Length of the latus rectum of the hyperbola $$ x y=c^{2}, $$ is
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$$\sqrt{2} {c}$$
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$$2c$$
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$$2\sqrt{2}{c}$$
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$$4c$$
Lissajous figure obtained by combining x=ASin $$\omega t$$ and y=ASin $$(\omega t+\Pi /4)$$ will be
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an ellipse
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a straight line
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a circle
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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
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$$\frac{{2a}}{3}$$
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a
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$$\frac{{4a}}{3}$$
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$$\frac{{5a}}{3}$$
Vertex of the parabola $$2{ y }^{ 2 }+3y+4x-1=0$$ is
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$$\left( \dfrac { 25 }{ 32 } ,\dfrac { -7 }{ 4 } \right) $$
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$$\left( \dfrac { 25 }{ 32 } ,\dfrac { -3 }{ 4 } \right) $$
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$$\left( \dfrac { 15 }{ 32 } ,\dfrac { 7 }{ 4 } \right) $$
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$$\left( \dfrac { 17 }{ 32 } ,\dfrac { -3 }{ 4 } \right) $$
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
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$$\dfrac {3}{2}$$
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$$\dfrac {\sqrt {5}}{2}$$
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$$\dfrac {3\sqrt {5}}{2}$$
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$$\dfrac {5}{2}$$
If the radius of the circle $${x}^{2}+{y}^{2}+2gx+2fy+c=0$$ be $$r$$, then it will touch both the axes, if
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$$g=f=c$$
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$$g=f=c=r$$
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$$g=f=\sqrt{c}=r$$
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$$g=f$$ 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
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$$(2,6)$$
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$$(1,4)$$
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$$(0,2)$$
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$$(0,4)$$
The equations $$x=\frac{t}{4},y=\frac{t^2}{4}$$ represents
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An ellipse
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A parabola
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A circle
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A hyperbola
If the radius of the circle $$ x ^ { 2 } + y ^ { 2 } $$ - 18 x - 12 y + k = 0 be 11 then k =
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347
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4
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-4
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97
The latus rectum of a parabola whose focal chord is PSQ such that SP=3 and SQ=2 is given by
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24/5
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12/5
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6/5
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23/5
If $$\frac { (3x-4y-{ 1) }^{ 2 } }{ 100 } -\frac { (4x+3y-{ 1) }^{ 2 } }{ 225 } =1,$$ then
length latusrectum of hyperbola is-
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$$\frac { 9 }{ 2 } $$
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$$\frac { 40 }{ 3 } $$
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9
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$$\frac { 8 }{ 3 } $$
If $$y ^ { 2 } - 2 x - 2 y + 5 = 0$$ is
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a circle with centre $$( 1,1 )$$
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a parabola with vertex
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a parabola with directrix $$x = 3 / 2$$
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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
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$$9$$
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$$\sqrt { 17 } $$
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$$8$$
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None of these.
The equation $$\dfrac{x^2}{10 - a} + \dfrac{y^2}{4 - a} = 1$$ , represents an ellipse , if
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$$a < 4$$
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$$a > 4$$
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$$4 < a < 10$$
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$$a > 10$$
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 :
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$$2$$
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$$4$$
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$$2 \sqrt{3}$$
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$$2 \sqrt{2}$$
$$C:\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1$$
The equation of parabolas with same latus- rectum as conic C, is/are
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$${y^2} - 5x + 3 = 0$$
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$${y^2} +6x - 21 = 0$$
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$${y^2} - 6x - 21 = 0$$
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$${y^2} +6x + 3 = 0$$
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
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$$
3 x ^ { 2 } - y ^ { 2 } = 3
$$
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$$
x ^ { 2 } - 3 y ^ { 2 } = 3
$$
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$$
3 x ^ { 2 } - y ^ { 2 } = 9
$$
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$$
x ^ { 2 } - 3 y ^ { 2 } = 9
$$
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
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15
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14
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13
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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
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0%
4
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2
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1/5
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2/5
equation of latus rectum of the parabola $${ y }^{ 2 }-16x-6y+1=0$$ is
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x-7=0
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x+7=0
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2x-7=0
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2x+7=0
The latus rectum of a parabola whose directrix is x + y- 2 =0 and focus is (3, -4) is
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$$-3\sqrt { 2 } $$
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$$3\sqrt { 2 } $$
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$$2\sqrt { 2 } $$
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$$3\sqrt { 2 } $$
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
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$$x^2+y^2-8x+12=0$$
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$$x^2+y^2-8x-12=0$$
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$$x^2+y^2-8x+9=0$$
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$$x^2+y^2=4$$
If e,e' be the eccentricities of two conics S and S' and if $$e^2+e^{,2}=3$$, then bothe S and S'
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Ellipses
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Parabola
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Hyperbola
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None of these
The equation of circle with centre (1, 2) and tangent $$x + y - 5 = 0$$ is
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$${x^2} + {y^2} + 2x - 4y + 6 = 0$$
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$${x^2} + {y^2} - 2x - 4y + 3 = 0$$
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$${x^2} + {y^2} - 2x + 4y + 8 = 0$$
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$${x^2} + {y^2} - 2x - 4y + 8 = 0$$
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
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$$\frac{x^2}{50}+\frac{y^2}{13}=1$$
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$$\frac{x^2}{52}+\frac{y^2}{13}=1$$
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$$\frac{x^2}{13}+\frac{y^2}{52}=1$$
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$$\frac{x^2}{52}+\frac{y^2}{17}=1$$
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 $$(\alpha ,\beta )$$
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(1, 3)
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(1, 5)
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(-5, 1)
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(-5, -1)
The equation of a circle with origin as a centre and passing through equilateral whose median is of length 3a is
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$${ x }^{ 2 }+{ y }^{ 2 }={ 9a }^{ 2 }$$
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$${ x }^{ 2 }+{ y }^{ 2 }={ 16a }^{ 2 }$$
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$${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$
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None of these
If the parabola $${ y }^{ 2 }=4ax$$ passes through (2,6) then the equation of the latusrectum is
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2x -9 = 0
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4x + 9 = 0
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2x + 9 = 0
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4x - 9 = 0
The length of the latus rectum of the parabola $$4 y ^ { 2 } + 2 x - 20 y + 17 = 0$$ is
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$$3$$
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$$6$$
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$$1 / 2$$
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None
The length of the latus rectum of the parabola $${ y }^{ 2 }-4x+4y+8=0\quad is$$
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8
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6
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4
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2
The equation $$\dfrac { { x }^{ 2 } }{ 12-a } +\dfrac { { y }^{ 2 } }{ 4-a } =1$$ represent an ellipse, if:
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$$a < 12$$
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$$ a < 4$$
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$$ a > 12$$
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$$ a > 4$$
The equation of circles passing through (3,-6) touching both the axes is
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$$x^{2}+y^{2}-6x+6y+9=0$$
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$$x^{2}+y^{2}+6x-6y+9=0$$
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$$x^{2}+y^{2}-30x-30y+225=0$$
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$$x^{2}+y^{2}-30x+30y+225=0$$
If $$a\neq b$$ the parametric equations $$x=a(cos \Theta +sin \Theta ),y=b(cos\Theta -sin\Theta )$$ represents
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Hyperbola
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A circle
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An ellipse
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A pair of straight lines
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A parabola
A common tangent to the conics $${ x }^{ 2 }=6y$$ and $$2{ x }^{ 2 }-4{ y }^{ 2 }=9$$, is __________.
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$$x+y=1$$
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$$x-y=1$$
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$$x+y=\dfrac { 9 }{ 2 }$$
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$$x-y=\dfrac { 3 }{ 2 } $$
The latus rectum of a parabola whose focal chord is PSQ such that SP=3 and SQ=2 is given by
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24/5
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12/5
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6/5
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none of these
The eccentricity of the hyperbola $$x^{2}-3y^{2}+1=0$$ is
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$$\frac{1}{2}$$
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1
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2
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3
If $$5x+9=0$$ is the directrix of the hyperbola $$16{x}^{2}-9{y}^{2}=144$$, then its correponding focus is:
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$$\left( -\cfrac { 5 }{ 3 } ,0 \right) $$
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$$(5,0)$$
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$$(-5,0)$$
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$$\left( \cfrac { 5 }{ 3 } ,0 \right) $$
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
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$$x^{2}+y^{2}-2x-2y-3=0$$
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$$x^{2}+y^{2}+2x-2y-3=0$$
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$$x^{2}+y^{2}-4x-6y+9=0$$
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$$x^{2}+y^{2}+x-y+5=0$$
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
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$$3x^2 + 4y^2$$ = 1
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$$3x^2 + 4y^2$$ = 12
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$$4x^2 + 3y^2$$ = 12
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$$4x^2 + 3y^2$$ = 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
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$$\tan^{-1}(1)$$
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$$\cot^{-1}1 + \cot^{-1}\dfrac {1}{2} + \cot^{-1} \dfrac {1}{3}$$
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$$\tan^{-1} (\sqrt {3})$$
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$$\tan^{-1} (2) +\tan^{-1} (3)$$
Equation of a circle whose centre is in $$I$$ quadrant as $$\left(\alpha,\ \beta\right)$$ and touches $$x-$$axis will be:
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$$x^{2}+y^{2}-2\alpha x - 2\beta y + \alpha ^{2}$$
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$$x^{2}+y^{2}+2\alpha x - 2\beta y + \alpha ^{2}$$
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$$x^{2}+y^{2}-2\alpha x + 2\beta y + \alpha ^{2}$$
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$$x^{2}+y^{2} + 2\alpha x + 2\beta y + \alpha ^{2}$$
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
$$x^2+y^2-2\alpha x-2\beta y+\alpha^2=0$$
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