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CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 11 - MCQExams.com
CBSE
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?
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Locus of P is pair of lines
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Locus of P is circle
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Area bounded by locus of P and x-axis above the x-axis is $$50\pi$$
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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
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$$2$$
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$$4$$
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$$8$$
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$$16$$
If $${ e }_{ 1 }$$ 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 $${ e }_{ 1 }{ e }_{ 2 }=1$, then equation of the hyperbola is ___________________.
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$$\dfrac { { x }^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 16 } =1$$
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$$\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$$
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$$\dfrac { { x }^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 25 } =1$$
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none of these
Length of latus rectum of the parabola $$9x^{2}+16y^{2}+24xy-4x+3y=0$$ is :
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$$\dfrac{1}{20}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{5}$$
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$$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$$.
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$$x^2-y^2+2x+4y+20=0$$
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$$x^2+y^2+2x-4y+20=0$$
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$$x^2+y^2-2x+4y-20=0$$
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$$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$$ is
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$$x^{2}+y^{2}-4x+4y-1= 0$$
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$$x^{2}+y^{2}+2x+2y-1= 0$$
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$$x^{2}+y^{2}+2x-2y-1= 0$$
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$$x^{2}+y^{2}-2x+4y-1= 0$$
The equation of the circle passing through $$(4,6)$$ and having centre $$(1,2)$$ is
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$${x}^{2}+{y}^{2}-2x-4y-20=0$$
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$${x}^{2}+{y}^{2}-2x+4y-20=0$$
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$${x}^{2}+{y}^{2}+2x-4y-20=0$$
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$${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]=?$$
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$$5$$
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$$2$$
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$$3$$
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$$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$$ is-
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$$\frac{1}{2}$$
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2
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$$\frac{3}{2}$$
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$$\frac{5}{2}$$
If the line $$3x+4y=1$$ touches the parabola $$y^{2}=4ax$$ then length of its latus rectum is equal to
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$$3/16$$
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$$3/8$$
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$$3/4$$
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$$9/16$$
The parabola $$y^{6}=4ax$$ passes through the point $$(2,-6)$$, then the length of its latus rectum is
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$$18$$
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$$9$$
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$$6$$
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$$16$$
The length of the latus rectum of the parabola $$x^2=ay+by+c$$ is
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$$\dfrac{a}{4}$$
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$$\dfrac{a}{3}$$
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$$\dfrac{1}{a}$$
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$$\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}$$ square units, with coordinate axes. Its eccentricity is equal to
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$$4$$
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$$5$$
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$$\dfrac{9}{2}$$
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$$\sqrt{17}$$
In the line $$2x+3y=1$$ touches the parabola $$y^{2}=4a(x+a)$$, then the length of its latus rectum is
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$$\dfrac {8}{9}$$
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$$\dfrac {8}{13}$$
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$$4$$
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$$\dfrac {4}{9}$$
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$$\dfrac {4}{13}$$
If the latus rectum subtends a right angel at the center of a hyperbola, then its eccentricity is
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$$\sqrt{2}$$
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$$\dfrac{\sqrt{3}+1}{2}$$
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$$\dfrac{\sqrt{5}-1}{2}$$
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$$\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$$
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Circle with radius $$2$$ units
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Parabola
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A point circle
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Ellipse
If PQ is a double ordinate of the hyperbola $$\frac { { x }^{ 2 } }{ { a }^{ 2 } } \frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$$, such that OPQ is an equilateral tnangle, O being the centre of the hyperbola, then the eccentncity e of the hyperbola satisfies
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$$e=\frac { 2 }{ \sqrt { 3 } } $$
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$$e=\frac { \sqrt { 3 } }{ 2 } $$
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$$e>\frac { 2 }{ \sqrt { 3 } } $$
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1
A tangent to the curve y = f(x) cuts the line y = x at a point which is at a distance of 1 unit from y - axis. The equation of the curve is
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$$\dfrac{x - 1}{y - 1} = C$$
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$$\dfrac{x}{y} = C$$
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$$xy = C$$
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none of these
The length of the latus rectum of the parabola $$x^2-4x-8y+12=0$$ is
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4
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6
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8
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2
Eccentricity of the hyperbola whose asymptotes are given by $$3x + 2y +5 = 0$$ and $$2x +3y+ 5 = 0$$ is
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$$\sqrt 2 $$
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$$\dfrac{3}{2}$$
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$$2$$
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None of these
Explanation
$$\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 } $$
If $$x=2+3\cos\theta$$ and $$y=1-3\sin\theta$$ represent a circle then the centre and radius is?
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$$(2, 1), 9$$
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$$(2, 1), 3$$
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$$(1, 2), \dfrac{1}{3}$$
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$$(-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$$, is
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Parabola
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Ellipse
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Hyperbola
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Pair of lines
The graph of the conic $$x ^ { 2 } - ( y - 1 ) ^ { 2 } = 1$$ has one tangent line with positive slope that passes through the origin, the point of tangency being (a, b). Then Length of the latus rectum of the conic is
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$$1$$
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$$\sqrt { 2 }$$
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$$2$$
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None
Let PQ be a variable focal chord of the parabola $${ y }^{ 2 }=4ax$$ where vertex is A. Locus of, centre triangle APQ is a parabola $$'{ P }_{ 1 }'$$ Latus rectum of parabola $${ P }_{ 1 }$$ is
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$$\dfrac { 2a }{ 3 } $$
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$$\dfrac { 4a }{ 3 } $$
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$$\dfrac { 8a }{ 3 } $$
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$$\dfrac { 16a }{ 3 } $$
If S be the focus of a parabola and PQ be the focal chord,such that SP=3 and SQ = 6,then the length of latus rectum of the parabola is
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4
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2
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8
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16
The equation of the image of the circle $${ \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$$
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$${ \left( x-14 \right) }^{ 2 }\quad +\quad { \left( y-13 \right) }^{ 2 }\quad =1$$
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$${ \left( x-15 \right) }^{ 2 }\quad +\quad { \left( y-14 \right) }^{ 2 }\quad =1$$
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$${ \left( x-16 \right) }^{ 2 }\quad +\quad { \left( y-15 \right) }^{ 2 }\quad =1$$
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$${ \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)$$. Find the possible values of length of latus rectum
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$$(3,\infty)$$
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$$(1,3/2)$$
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$$(2,3)$$
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$$(-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$$ is greater than 2, then the length of its latus rectum lies in the interval:
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$$(1,3/2]$$
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$$(3/2,2]$$
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$$\left( 3,\infty \right) $$
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$$(2,3]$$
The lines $$2x-3y-5=0$$ and $$3x-4y=7$$ are diameters of a circle of area $$154 \ sq\ units$$, then the equation if the circle is :
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$$x^2 +y^2 +2x-2y-62=0$$
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$$x^2 +y^2 +2x-2y-47=0$$
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$$x^2 +y^2 -2x+2y-62=0$$
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$$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$$.
Axes are coordinates axes, the ellipse passes through the points where the straight line $$\dfrac {x}{4}+\dfrac {y}{3}=1$$ meets the coordinates axes. Then equation of the ellipses is
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$$\dfrac {x^{2}}{16}+\dfrac {y^{2}}{9}=1$$
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$$\dfrac {x^{2}}{64}+\dfrac {y^{2}}{36}=1$$
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$$\dfrac {x^{2}}{4}+\dfrac {y^{2}}{3}=1$$
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$$\dfrac {x^{2}}{8}+\dfrac {y^{2}}{6}=1$$
The length of the latus rectum of the parabola, $$y^2-6y+5x=0$$ is
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$$1$$
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$$3$$
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$$5$$
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$$7$$
The equation $${ x }^{ 2 }+{ y }^{ 2 }+4x+6y+13=0\quad $$ represents
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Circle
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Pair of coincident straight lines
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Pair of concurrent straight lines
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Point
Length of the latus rectum of the parabola $$25[(x-2)^{2}+(y-3)^{2}]=(3x-4y+7)^{2}$$ is:
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$$4$$
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$$2$$
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$$1/5$$
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$$2/5$$
Which of the following can be the equation of an ellipse?
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$$x^{2} + y^{2} = 5$$
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$$\dfrac {x^{2}}{9} + \dfrac {x^{2}}{9} = 1$$
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$$2x^{2} + 3y^{2} = 5$$
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$$2x + 2y = 5$$
The curve for which the normal at any point (x,y) and the line joining origin to that point form and isosceles triangle with the x-axis as base is
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an ellipse
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a rectangular hyperbola
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a circle
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circle.
The equation of circle at center $$(3,4)$$ and radius $$5$$.
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$$x^2+y^2-6x-8y=0$$
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$$x^2+y^2+3x+4y+5=0$$
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$$x^2+y^2+6x+8y=0$$
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None.
Explanation
The center of equation $$(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$$ passing through (a , 2a) are bisected on the line x + y =1, then the length of the latnsrectum can be
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2
0%
1
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4
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3
If two distinct chords of a parabola $${ y }^{ 2 }=4ax$$ passing through $$(a, 2a)$$ are bisected on the $$x+y=1$$, then the length of the latusrectum can be
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0%
2
0%
1
0%
4
0%
3
If the parabola $${y^2} = 4\,\,ax$$ passes through the point $$\left( { - 3,2} \right),$$ then the length of its latus rectum is
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2 / 3
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4 / 3
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1 / 3
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4
Match the column I with column II and mark the correct option from the codes given below.
Column I
Column II
i
ii
iii
iv
v
$$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$$
p
q
r
s
Ellipse
Hyperbola
Parabola
Pair of straight lines
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(i)$$=$$s,(ii)$$=$$q,(iii)$$=$$s,(iv)$$=$$p,(v)$$=$$s
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(i)$$=$$p,(ii)$$=$$r,(iii)$$=$$q,(iv)$$=$$r,(v)$$=$$s
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(i)$$=$$q,(ii)$$=$$p,(iii)$$=$$r,(iv)$$=$$s,(v)$$=$$q
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(i)$$=$$r,(ii)$$=$$s,(iii)$$=$$p,(iv)$$=$$q,(v)$$=$$r
If $$ P S Q $$ is the focal chord of the parabola $$ y^{2}=8 x $$ such that $$ S P=6 $$ . Then the length of $$SQ$$ is
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$$6$$
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$$4$$
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$$3$$
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none of these
If $$x ^ { 2 } + y ^ { 2 } - 2 x + 2 a y + a + 3 = 0$$ represents a real circle with non-zero radius, then most appropriate is?
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$$a \in ( - \infty , - 1 )$$
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$$a \epsilon ( - 1,2 )$$
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$$a \epsilon ( 2 , \infty )$$
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$$a \epsilon ( - \infty , - 1 ) \cup ( 2 , \infty )$$
Equation of the circle whose radius is 3 and centre is $$(-1,2)$$
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$${ x }^{ 2 }+{ y }^{ 2 }-2x+4y=4$$
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$${ x }^{ 2 }+{ y }^{ 2 }+2x-4y+4=0$$
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$${ x }^{ 2 }+{ y }^{ 2 }+2x-4y=4$$
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$${ 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}=$$
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$$\dfrac { 13 }{ 36 } $$
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$$\dfrac { 36 }{ 13 } $$
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$$\dfrac { 16 }{ 33 } $$
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$$\dfrac { 33 }{ 16 } $$
If the equation $$p{ x }^{ 2 }+(2-q)xy+3{ y }^{ 2 }-6qx+30y+6q=0$$ represents a circle, then the values of p and q are
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3, 1
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2, 2
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3, 2
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3, 4
The length of the lotus rectum of the parabola $$9x^{2}-6x+36y+19=0$$
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$$36$$
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$$9$$
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$$6$$
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$$4$$
Equation of the circle whose radius is $$a+b$$ and centre $$(a, -b)$$
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$${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=2ab$$
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$${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=-2ab$$
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$${ x }^{ 2 }+{ y }^{ 2 }-2ax+2by=ab$$
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$${ 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.$$ Find the equation of the hyperbola, if its eccentricity is 2.
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$$\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1$$
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$$\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =2$$
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$$\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 12 } =1$$
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None of these
If $$ x^{2}+y^{2}-2 x+2 a y+a+3=0 $$ represents a real circle with non-zero radius, then most appropriate is
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$$
a \in(-\infty,-0)
$$
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$$
a \in(-1,2)
$$
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$$
a \in(2,\infty)
$$
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$$
a \in(-\infty,-1) \cup(2 \infty)
$$
Find the equation of a circle whose cemtre is $$\left( {2,1} \right)$$ and radius is 3
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$${x^2} + {y^2} + 4x - 2y + 4 = 0$$
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$${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$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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