MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 10

Find the length of latus rectum of the parabola
$$(a^{2}+b^{2})(x^{2}+y^{2})=(bx+ay-ab)^{2}$$

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$$\displaystyle \frac{ab}{\sqrt{a^{2}+b^{2}}}$$
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$$\displaystyle \frac{2ab}{\sqrt{a^{2}+b^{2}}}$$
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$$a+b$$
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$$a+b+\dfrac{ab}{\sqrt{ab}}$$

If a hyperbola passes through the foci of the ellipse $$\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$$ and its traverse and conjugate axis coincide with major and minor axes of the ellipse, and product of the eccentricities is 1, then:

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Equations of the hyperbola is $$\displaystyle \frac {x^2}{9} - \frac {y^2}{16} = 1$$
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Equations of the hyperbola is $$\displaystyle \frac {x^2}{9} - \frac {y^2}{25} = 1$$
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Focus of the hyperbola is $$\displaystyle (5, 0)$$
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Focus of the hyperbola is $$\displaystyle (5 \sqrt 3, 0)$$

The locus of the point $$(h,k)$$, if the point $$(\sqrt{3h}, \sqrt{3k + 2})$$ lies on the line $$x - y - 1 = 0$$, is a ?

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straight line
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circle
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parabola
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none of these

A circle touches the $$x$$-axis and also touches the circle with centre $$(0, 3)$$ and radius $$2$$. The locus of the centre of the circle is -

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a circle
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an ellipse
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a parabola
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a hyperbola

A point $$(\alpha, \beta)$$ lies on a circle $$x^2+y^2=1$$, then locus of the point $$(3\alpha +2\beta)$$ is a$$/$$an.

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Straight line
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Ellipse
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Parabola
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None of these

Find the equation of the circle that passes through the points $$(0,6),(0,0)$$ and $$(8,0)$$

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$${(x-4)}^{2}+{(y-3)}^{2}=25$$
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$$(x+4)^2+(y+3)^2=25$$
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$$(x-3)^2+(y-4)^2=25$$
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$$(x-4)^2+(y-3)^2=36$$

A circle and a parabola intersect at four points $$(x_1 , y_1), (x_2 , y_2), (x_3 , y_3)$$ and $$(x_4 , y_4)$$. Then $$y_1 + y_2 + y_3 + y_4$$ is equal to

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$$4$$
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$$3/2$$
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$$ 2$$
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$$0$$

A Iight ray gets reflected

from the $$x = -2$$. If the reflected touches the circle $$x^2 + y^2 = 4$$ and point of incident is $$(-2,-4)$$, then equation of incident ray is

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$$4y+3x+22=0$$
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$$3y+4x+20=0$$
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$$4y+2x+20=0$$
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$$y+x+6=0$$

Find the equation of the circle with center at $$(-3,5)$$ and passes through the point $$(5,-1)$$

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$${(x+3)}^{2}+{(y-5)}^{2}=100$$
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$$(x-3)^2+(y-5)^2$$
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$$(x+3)^2+(y-5)^2$$
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None of the above

Normals $$AA,A{A}_{1}$$ and $$A{A}_{2}$$ are drawn to the parabola $${ y }^{ 2 }=8x$$ from the point $$A(h,0)$$. If triangle $$O{A}_{1}{A}_{2}$$ is equilateral , then the possible value of $$h$$ is

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$$26$$
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$$24$$
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$$28$$
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None of these

Consider a parabola $$P$$ which touches $$y = 0$$ at $$(1, 0)$$ and $$x = 0$$ at $$(0, 2)$$, then latus rectum of $$P$$ is,

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$$\dfrac {9}{5\sqrt {5}}$$
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$$\dfrac {16\sqrt {5}}{5}$$
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$$\dfrac {16}{25}$$
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$$\dfrac {8}{25}$$

A circle touches the y-axis at $$(0, 2)$$ and has an intercept of $$4$$ units on the positive side of the x-axis. Then the equation of the circle is?

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$$x^2+y^2-4(\sqrt{2}x+y)+4=0$$
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$$x^2+y^2-4(x+\sqrt{2}y)+4=0$$
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$$x^2+y^2-2(\sqrt{2}x+y)+4=0$$
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none of these

Let $$S$$ be the focus of $${ y }^{ 2 }=4x$$ and a point $$P$$ be moving on the curve such that its abscissa is increasing at the rate of $$4 units/s$$. Then the rate of increase of the projection of $$SP$$ on $$x+y=1$$ when $$P$$ is at $$(4,4)$$ is

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$$\sqrt { 2 } $$
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$$-1$$
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$$-\sqrt { 2 } $$
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$$-3\sqrt { 2 } $$

O is the centre of a circle of diameter $$4$$cm and OABC is a square, if the shaded area is $$\displaystyle\frac{1}{3}$$ area of the square, then the side of the square is __________.

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$$\pi\sqrt{3}$$cm
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$$\sqrt{3\pi}$$cm
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$$3\sqrt{\pi}$$cm
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$$3\pi$$cm

The ratio of the areas of the triangle $$PQS$$ and $$PQR$$ is :

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$$1 : \sqrt2$$
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$$1 : 2$$
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$$1 : 4$$
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$$1 : 8$$

Explanation

Equation of parabola is $$y^2=8x$$

Equation of circle is $$x^2+y^2=9$$

To find point of intersection of both, put equation of parabola in equation of circle.

$$\therefore x^2+8x=9$$

$$\therefore x^{ 2 }+8x-9=0$$

$$\therefore x^{ 2 }+9x-x-9=0$$

$$\therefore x\left( x+9 \right) -1\left( x+9 \right) =0$$

$$\therefore \left( x+9 \right) \left( x-1 \right) =0$$

$$\therefore x=-9$$ and $$x=1$$

But, $$x\neq -9$$ as $$y^2$$ cannot be negative

$$\therefore x=1$$

From equation of parabola,

$${ y }^{ 2 }=8$$

$$\therefore { y }=\pm 2\sqrt { 2 }$$

Thus, points of intersection are $$P\left( 1,2\sqrt { 2 } \right)$$ and $$Q\left( 1,-2\sqrt { 2 } \right)$$

Consider tangent to the circle which intersects x axis at point R.

Equation of tangent is,

$$x{ x }_{ 1 }+y{ y }_{ 1 }=9$$

$$\therefore x\left( 1 \right) +y\left( 2\sqrt { 2 } \right) =9$$

$$\therefore x+2\sqrt { 2 } y=9$$

When $$y=0$$, $$x=9$$

Thus, coordinates of point R are $$R\left( 9,0 \right)$$

Consider tangent to the parabola which intersects x axis at point S.

Equation of tangent is,

$$y{ y }_{ 1 }=4a\left( x+{ x }_{ 1 } \right)$$

$$\therefore y\left( 2\sqrt { 2 } \right) =8\left( x+1 \right)$$

When $$y=0$$, $$x=-1$$

Thus, coordinates of point S are $$R\left( -1,0 \right)$$

Area of triangle PQS is,

$$A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times PQ\times SM$$

$$\therefore A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 2$$

$$\therefore A\left( \triangle PQS \right) =4\sqrt { 2 }$$

Area of triangle PQR is,

$$A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times PQ\times RM$$

$$\therefore A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 8$$

$$\therefore A\left( \triangle PQR \right) =16\sqrt { 2 }$$

$$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =\dfrac { 4\sqrt { 2 } }{ 16\sqrt { 2 } }$$

$$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =1:4$$

1973247_874922_ans_2efa3d16990d43a3ba04c0ccb23185cd.png

If eccentricity of both ellipses are same, then their eccentricity is

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$$2-\sqrt { 3 } $$
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$$\sqrt { 2 } -1$$
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$$\cfrac { 3-\sqrt { 5 } }{ 2 } $$
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$$\cfrac { \sqrt { 5 } -1 }{ 2 } \quad $$

An endless inextensible string of length $$15$$m passes around the pins, A & B which are $$5$$m apart. This string is always kept tight and a small ring, R of negligible dimensions, inserted in this string is made to move in a path keeping all segments RA, AB, RB tight (as mentioned earlier). The ring traces a path, given by conic C, then.

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Conic C is an ellipse with eccentricity $$\displaystyle\frac{1}{2}$$
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Conic C is an hyperbola with eccentricity $$2$$
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Conic C is an ellipse with eccentricity $$\displaystyle\frac{2}{3}$$
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Conic C is a hyperbola with eccentricity $$\displaystyle\frac{3}{2}$$

$$S_{1}$$ and $$S_{2}$$ are the foci of an ellipse of major axis of length 10 units, and P is any point on the ellipse such that the perimeter or triangle $$PS_{1}S_{2}$$ isThen the eccentricity of the ellipse is

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0.5
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0.25
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0.28
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0.75

$$Center\quad of\quad the\quad hyperbola\quad { x }^{ 2 }+4{ y }^{ 2 }+6xy+8x-2y+7=0\quad is\quad $$

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$$(1,1)$$
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$$(0,2)$$
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$$(2,0)$$
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$$None\quad of\quad these$$

'O' is the vertex of the parabola $${ y }^{ 2 }=8x$$ and L is the upper end of the latus rectum. If LH is drawn perpendicular to OL meeting OX in H, then the length of the double ordinate through H is $$\lambda \sqrt { 5 } $$ where $$\lambda $$ is equal to

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$$2$$
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$$4$$
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$$6$$
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$$8$$

Which of the following equations does not represent a hyperbola?

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$$xy = 4$$
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$$
\dfrac{1}
{x^2} + \dfrac{1}
{y^2} = \dfrac{1}
{4}
$$
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$$
x^2 - xy + y^2 = 4
$$
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$$
x^2 - 4xy + 3y^2 = 1
$$

The set of points $$(x, y)$$ whose distance from the line $$y = 2x + 2$$ is the same as the distance from $$(2, 0)$$ is a parabola. This parabola is congruent to the parabola in standard form $$y = Kx^{2}$$ for some $$K$$ which is equal to

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$$\dfrac {\sqrt {5}}{12}$$
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$$\dfrac {\sqrt {5}}{4}$$
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$$\dfrac {4}{\sqrt {5}}$$
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$$\dfrac {12}{\sqrt {5}}$$

From the point $$A\left(0,3\right)$$ on the circle $${x}^{2}-4x+{\left(y-3\right)}^{2}=0$$ a chord $$AB$$ is drawn and extended to a point $$M$$ such that $$AM=2AB$$.The locus is

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$${x}^{2}+{y}^{2}-8x-6y+9=0$$
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$${x}^{2}+{y}^{2}+8x-6y+9=0$$
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$${x}^{2}+{y}^{2}-8x+6y+9=0$$
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$${x}^{2}+{y}^{2}+8x+6y+9=0$$

The line $$2x + y = 1$$ is tangent to the hyperbola $$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$. If this line passes through the point of intersection of the nearest directrix and the x-axis, then eccentricity of the hyperbola.

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1
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2
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3
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4

The equation of the circle having the lines $$x^2 + 2xy + 3x + 6y = 0$$ as its normals and having size just sufficient to contain the circle $$x (x - 4) + y(y - 3) = 0$$ is

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$$x^2 + y^2 + 3x - 6y - 40 = 0$$
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$$x^2 + y^2 + 6x - 3y - 45 = 0$$
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$$x^2 + y^2 + 8x + 4y - 20 = 0$$
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$$x^2 + y^2 + 4x + 8y + 20 =0$$

$$LL^1$$ is the latus rectum of an ellipse and $$\Delta S^1LL^1$$ is an equilateral triangle. Then $$e=?$$

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$$\dfrac{1}{\sqrt{2}}$$
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$$\dfrac{1}{\sqrt{3}}$$
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$$\dfrac{1}{\sqrt{5}}$$
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$$\sqrt{\dfrac{2}{3}}$$

Explanation

Let equation of the ellipse is $$\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ b^{ 2 } } =1$$

Refer the figure. By distance formula,

$${ S }^{ ' }L=\sqrt { { \left( ae-\left( -ae \right) \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } \right) }^{ 2 } }$$

$$\therefore { S }^{ ' }L=\sqrt { { \left( 2ae \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } \right) }^{ 2 } }$$

$$\therefore { S }^{ ' }L=\sqrt { 4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } } }$$

$$\therefore { \left( { S }^{ ' }L \right) }^{ 2 }=4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } }$$ (1)

Similarly, $$L{ L }^{ ' }=\sqrt { { \left( ae-ae \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } -\left( -\frac { { b }^{ 2 } }{ a } \right) \right) }^{ 2 } }$$

$$\therefore L{ L }^{ ' }=\sqrt { { \left( \frac { { b }^{ 2 } }{ a } +\frac { { b }^{ 2 } }{ a } \right) }^{ 2 } } $$

$$\therefore L{ L }^{ ' }=\sqrt { { \left( \frac { 2{ b }^{ 2 } }{ a } \right) }^{ 2 } } $$

$$\therefore L{ L }^{ ' }=\sqrt { { \frac { 4{ b }^{ 4 } }{ { a }^{ 2 } } } } $$

$$\therefore { \left( L{ L }^{ ' } \right) }^{ 2 }=\frac { 4{ b }^{ 4 } }{ { a }^{ 2 } } $$ (2)

Now, given $$\triangle { S }^{ ' }LL^{ ' }$$ is equilateral triangle

$$\therefore { \left( { S }^{ ' }L \right) }^{ 2 }={ \left( L{ L }^{ ' } \right) }^{ 2 }$$

$$\therefore 4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } } =\frac { 4{ b }^{ 4 } }{ { a }^{ 2 } }$$

$$\therefore 4{ a }^{ 2 }{ e }^{ 2 }=\frac { 3{ b }^{ 4 } }{ { a }^{ 2 } } $$

$$\therefore 4{ e }^{ 2 }=\frac { 3{ b }^{ 4 } }{ { a }^{ 4 } } $$ (3)

Now, $$\frac { { b }^{ 2 } }{ { a }^{ 2 } } =1-{ e }^{ 2 }$$

$$\therefore { \left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) }^{ 2 }={ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$$

$$\therefore \frac { { b }^{ 4 } }{ { a }^{ 4 } } ={ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$$ (4)

From equations (3) and (4), we can write,

$$4{ e }^{ 2 }=3{ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$$

$$\therefore 4{ e }^{ 2 }=3\left( 1-2{ e }^{ 2 }+{ e }^{ 4 } \right)$$

$$\therefore 4{ e }^{ 2 }=3-6{ e }^{ 2 }+3{ e }^{ 4 }$$

$$\therefore 3{ e }^{ 4 }-10{ e }^{ 2 }+3=0$$

$$\therefore 3{ e }^{ 4 }-9{ e }^{ 2 }-{ e }^{ 2 }+3=0$$

$$\therefore 3{ e }^{ 2 }\left( { e }^{ 2 }-3 \right) -1\left( { e }^{ 2 }-3 \right) =0$$

$$\therefore \left( 3{ e }^{ 2 }-1 \right) \left( { e }^{ 2 }-3 \right) =0$$

$$\therefore { e }^{ 2 }=\frac { 1 }{ 3 } $$ or $${ e }^{ 2 }=3$$

$$\therefore { e }=\frac { 1 }{ \sqrt { 3 } } $$ or $${ e }=\sqrt { 3 }$$

But $${ e }<1$$

$$\therefore { e }\neq \sqrt { 3 } $$

$$\therefore { e }=\frac { 1 }{ \sqrt { 3 } } $$

1946415_1082283_ans_a6c838f567cd4a21a982152f00f47135.png

Circles are drawn on chords of the rectangular hyperbola $$xy=4$$ parallel to the line $$y=x$$ as diameters.All such circles pass through two fixed points whose coordinates are

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$$\left(2,2\right)$$
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$$\left(2,-2\right)$$
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$$\left(-2,2\right)$$
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$$\left(-2,-2\right)$$

Explanation

Given:Rectangular hyperbola $$xy=4={c}^{2}$$

$$\Rightarrow\,{c}^{2}=4$$

$$\Rightarrow\,c=2$$

Let $$P$$ and $$Q$$ be the end points on the Rectangular hyperbola where $$P\left(2{t}_{1},\dfrac{2}{{t}_{1}}\right)$$ and $$Q\left(2{t}_{2},\dfrac{2}{{t}_{2}}\right)$$

Using the end points of diameter the equation of the circle

$$C:\left(x-2{t}_{1}\right)\left(x-2{t}_{2}\right)+\left(y-\dfrac{2}{{t}_{1}}\right)\left(y-\dfrac{2}{{t}_{2}}\right)=1$$ ..........$$(1)$$

Now,Slope of the $$PQ=\dfrac{\dfrac{2}{{t}_{2}}-\dfrac{2}{{t}_{1}}}{2{t}_{2}-2{t}_{1}}$$

$$=\dfrac{2\left(\dfrac{1}{{t}_{2}}-\dfrac{1}{{t}_{1}}\right)}{2\left({t}_{2}-{t}_{1}\right)}$$

$$=\dfrac{\dfrac{{t}_{1}-{t}_{2}}{{t}_{1}{t}_{2}}}{\left({t}_{2}-{t}_{1}\right)}$$

$$=\dfrac{-1}{{t}_{1}{t}_{2}}$$

Hence $$PQ$$ is the diameter for circle and it is parallel to the line $$y=x$$

Slope of $$PQ=$$Slope of the line $$y=x$$

$$\Rightarrow\,\dfrac{-1}{{t}_{1}{t}_{2}}=1$$

$$\Rightarrow\,{t}_{1}{t}_{2}=-1$$

$$(1)\Rightarrow\,\left(x-2{t}_{1}\right)\left(x-2{t}_{2}\right)+\left(y-\dfrac{2}{{t}_{1}}\right)\left(y-\dfrac{2}{{t}_{2}}\right)=1$$

$$\Rightarrow\,x\left(x-2{t}_{2}\right)-2{t}_{1}\left(x-2{t}_{2}\right)+y\left(y-\dfrac{2}{{t}_{2}}\right)-\dfrac{2}{{t}_{1}}\left(y-\dfrac{2}{{t}_{2}}\right)=1$$

$$\Rightarrow\,{x}^{2}-2x{t}_{2}-2x{t}_{1}+4{t}_{1}{t}_{2}+{y}^{2}-\dfrac{2y}{{t}_{2}}-\dfrac{2y}{{t}_{1}}+\dfrac{4}{{t}_{1}{t}_{2}}=1$$

$$\Rightarrow\,{x}^{2}-2x{t}_{2}-2x{t}_{1}+4\times -1+{y}^{2}-\dfrac{2y}{{t}_{2}}-\dfrac{2y}{{t}_{1}}+\dfrac{4}{\times -1}=1$$ using $${t}_{1}{t}_{2}=-1$$

$$\Rightarrow\,{x}^{2}+{y}^{2}-2x\left({t}_{2}+{t}_{1}\right)-4-2y\left(\dfrac{1}{{t}_{2}}+\dfrac{1}{{t}_{1}}\right)-4=1$$

$$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t}_{2}+{t}_{1}\right)-2y\left(\dfrac{{t}_{1}+{t}_{2}}{{t}_{1}{t}_{2}}\right)=1$$

$$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t}_{2}+{t}_{1}\right)-2y\left(\dfrac{{t}_{1}+{t}_{2}}{-1}\right)=1$$ using $${t}_{1}{t}_{2}=-1$$

$$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t}_{2}+{t}_{1}\right)+2y\left({t}_{1}+{t}_{2}\right)=1$$

$$\Rightarrow\,{x}^{2}+{y}^{2}-8+\left(2y-2x\right)\left({t}_{2}+{t}_{1}\right)=1$$ is of the form $$C+\lambda\,L$$

where $$C={x}^{2}+{y}^{2}-8=0$$ is the equation of a circle.

and $$L=2y-2x=0$$ is the equation of a line.

$$\Rightarrow\,y-x=0$$ or $$x=y$$

Substituting $$x=y$$ in the equation $${x}^{2}+{y}^{2}-8=0$$ we get

$$\Rightarrow\,2{x}^{2}-8=0$$

$$\Rightarrow\,2\left({x}^{2}-4\right)=0$$

$$\Rightarrow\,\left(x-2\right)\left(x+2\right)=0$$

$$\therefore\,x=2,-2$$

$$\Rightarrow\,y=2,-2$$ since $$x=y$$

Hence the coordinates of the fixed points are $$\left(2,2\right)$$ and $$\left(-2,-2\right)$$

Consider
the set of hyperbola $$xy = {\text{ }}K,{\text{ K}} \in {\text{R,}}$$ let $${e_1}$$ be eccentricity
when $$K = \sqrt {2017} $$ and $${e_2}$$ be the
eccentricity when $$K = \sqrt {2018} $$ , then $${e_1} - {e_2}$$ is equal to

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-1
0%
0
0%
2
0%
1

The centre of a circle is $$(2, -3)$$ and the circumference is $$10\pi$$. Then, the equation of the circle is

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$${x}^{2}+{y}^{2}+4x+6y+12=0$$
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$${x}^{2}+{y}^{2}-4x+6y+12=0$$
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$${x}^{2}+{y}^{2}-4x+6y-12=0$$
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$${x}^{2}+{y}^{2}-4x-6y-12=0$$

A conic $$C$$ passes through the points $$(2,4)$$ and is such that the segment of any of its tangents at any point contained between the co-ordinate axis is biscected at the point of tangency. Let $$S$$ denotes circle described on the foci $${F_1}$$ and $${F_2}$$ of the conic $$C$$ as diameter.
Equation of the circle $$S$$ is

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$${x^2} + {y^2} = 16$$
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$${x^2} + {y^2} = 8$$
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$${x^2} + {y^2} = 32$$
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$${x^2} + {y^2} = 4$$

The centre of a circle passing through the point $$(0,0),(1,0)$$ and touching the circle $$x^{2}+y^{2}=9$$ is ?

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$$\left(\dfrac {3}{2},\dfrac {1}{2}\right)$$
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$$\left(\dfrac {1}{2},\dfrac {3}{2}\right)$$
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$$\left(\dfrac {1}{2},\dfrac {1}{2}\right)$$
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None of these

If the centroid of an equilateral triangle $$(1,1)$$ and its one vertex is $$(-1,2)$$ , then the equation of the circumcircle 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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$${x^2} + {y^2} + 2x + 2y - 3 = 0$$
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$${(x + 2)^2} + {y^2} = 5$$

The focus of extremities of the latus rectum of the family of the ellipse $${b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}{\text{ is }}\left( {b \in R} \right)$$

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$${x^2} - ay = {a^3}$$
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$${x^2} - ay - {e^2}$$
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$${x^2} \pm ay = {a^2}$$
0%
$${x^2} + ay - {b^2}$$

The equation of the circle touches y axis and having radius $$2$$ units and centre is $$(-2, -3)$$?

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$$x^2+y^2-4x-9y-4=0$$
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$$x^2+y^2+4x+9y+4=0$$
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$$x^2+y^2+4x+6y+9=0$$
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$$x^2+y^2-4x-6y-9=0$$

The normal at $$ P(8, 8)$$ to the parabola $$y^2=8x$$ cuts it again at Q then PQ =

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$$10$$
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$$10\sqrt5$$
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$$4\sqrt5$$
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$$50$$

Circles are drawn passing through the origin $$O$$ to intersect the coordinate axes at point $$P$$ and $$Q$$ such that $$m$$. $$PO+n.OQ=k$$, then the fixed point satisfy all of them, is given by

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$$(m,n)$$
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$$\dfrac{m^{2}}{k},\dfrac{n^{2}}{k}$$
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$$\dfrac{mk}{m^{2}+n^{2}},\dfrac{nk}{m^{2}+n^{2}}$$
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$$(k,k)$$

The vertex $$A$$ of the parabola $${y}^{2}=4ax$$ is joined to any point $$P$$ on it and $$PQ$$ is drawn at right angles to $$AP$$ to meet the axis in $$Q$$. Projection of $$PQ$$ on the axis is equal to

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twice the length of latus rectum
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the latus length of rectum
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half the length of latus rectum
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one fourth of the length of latus rectum

If equation $$(5x-1)^{2}+(5y-2)^{2}=(\lambda^{2}-2\lambda+1)(3x+4y-1)^{2}$$ represents an ellipse, then $$\lambda \in$$

0%
$$(0, 1)$$
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$$(0, 2)$$
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$$(1, 2)$$
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$$(0, 1)\cup (1, 2)$$

The locus of the mid points of the portion of the tangents to the ellipse intercepted between the axes

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$$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=4$$
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$$\dfrac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4$$
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$$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=4$$
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none of these

If the curves $$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1$$ and $$\dfrac{x^{2}}{l^{2}}-\dfrac{y^{2}}{4}=1$$ cut each other orthogonally then $$l^{2}=$$

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

Equation of the latus rectum of the hyperbola $$(10x - 5)^{2} + (10y - 2)^{2} = 9(3x + 4y - 7)^{2}$$ is

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$$y - 1/5 =-3/4(x - 1/2)$$
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$$x - 1/5 =-3/4(y - 1/2)$$
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$$y + 1/5 =-3/4(x + 1/2)$$
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$$x + 1/5 =-3/4(y + 1/2)$$

If the equation $$\frac { \lambda ( x + 1 ) ^ { 2 } } { 3 } + \frac { ( y + 2 ) ^ { 2 } } { 4 } = 1$$ represents a circle then $$\lambda =$$

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1
0%
$$\frac { 3 } { 4 }$$
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0
0%
$$- \frac { 3 } { 4 }$$

The area enclosed between the parabolas $$y^2 = 4x$$ and $$x^2 = 4y$$ is

0%
$$\dfrac{16}{3}$$ sq. unit
0%
$$\dfrac{3}{4}$$ sq. unit
0%
$$\dfrac{3}{16}$$ sq. unit
0%
None of these

Equation of the curve passing through the point $$(1,\ 2)$$ such that the intercept on the $$x-$$axis cut off between the tangent and origin is twice the abscissa is given by:

0%
$$xy=2$$
0%
$$xy=1$$
0%
$$xy=2y$$
0%
$$xy=2x$$

Let $$S$$ and $$S ^ { 1 }$$ are the foci of an ellipse whose eccentricity is $$\frac { 1 } { \sqrt { 2 } } , B$$ and $$B ^ { 1 }$$ are the ends of minor axis then $$S B S ^ { \prime } B ^ { 1 }$$ forms $$a$$ ________________.

0%
Parallelgram
0%
Rhombus
0%
Square
0%
Rectangle

$$D.E$$ of the curve for which the initial ordinates of any tangent is equal to the corresponding number

0%
Second degree in $$x$$
0%
Hemiohenous of second degree
0%
Has separable variables
0%
is of second degree

ABCD is a square of side 1 unit. A circle passes through vertices A,B of the square and the remaining two vertices of the square lie out side the circle. The length of the tangent draw to the circle from vertex D is 2 units. The radius of the circle is

0%
$$\sqrt{5}$$
0%
$$\dfrac{1}{2}$$$$\sqrt{10}$$
0%
$$\dfrac{1}{3}$$$$\sqrt{12}$$
0%
$$\sqrt{8}$$

The coordinates of the foci of the hyperbola $$xy=c^2$$ are

0%
$$(\pm C, \pm C)$$
0%
$$(\pm \dfrac{c}{\sqrt{2}}< \pm\dfrac{c}{\sqrt{2}})$$
0%
$$(\pm 2c, \pm 2c)$$
0%
$$(\pm \sqrt{2}c, \pm \sqrt{2}c)$$

The locus of point of intersection $$P$$ of tangents to ellipse $$2x^{2}+3y^{2}=6$$ at $$A$$ and $$B$$ if $$AB$$ subtend $$90^{o}$$ angle at centre of ellipse is an ellipse whose eccentricity is equal to

0%
$$\surd {5}/4$$
0%
$$\surd {5}/3$$
0%
$$2/\surd {5}$$
0%
$$none\ of\ these$$

The equation of the circle, passing through the point $$\left(2,8\right)$$, touching the lines $$4x-3y-24=0$$ and $$4x+3y-42=0$$ and having x coordinate of the centre of the circle numerically less then or equal to 8 is

0%
$$x^2+y^2+4x-6y-12=0$$
0%
$$x^2+y^2-4x+6y-12=0$$
0%
$$x^2+y^2-4x-6y-12=0$$
0%
None of these

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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