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1

MCQ (Single Correct Answer)

Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}} $$ is

A

$$5{x^2} + 3{y^2} - 48 = 0$$

B

$$3{x^2} + 5{y^2} - 15 = 0$$

C

$$5{x^2} + 3{y^2} - 32 = 0$$

D

$$3{x^2} + 5{y^2} - 32 = 0$$

Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$

Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$

$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$

So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$

It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$

Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$

$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$

So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$

2

MCQ (Single Correct Answer)

If two tangents drawn from a point $$P$$ to the parabola $${y^2} = 4x$$ are at right angles, then the locus of $$P$$ is

A

$$2x+1=0$$

B

$$x=-1$$

C

$$2x-1=0$$

D

$$x=1$$

The locus of perpendicular tangents is directrix

i.e., $$x=-1$$

i.e., $$x=-1$$

3

MCQ (Single Correct Answer)

The ellipse $${x^2} + 4{y^2} = 4$$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $$(4,0)$$. Then the equation of the ellipse is :

A

$${x^2} + 12{y^2} = 16$$

B

$$4{x^2} + 48{y^2} = 48$$

C

$$4{x^2} + 64{y^2} = 48$$

D

$${x^2} + 16{y^2} = 16$$

The given ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$

So $$A=(2,0)$$ and $$B = \left( {0,1} \right)$$

If $$PQRS$$ is the rectangular in which it is inscribed, then

$$P = \left( {2,1} \right).$$

Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

be the ellipse circumscribing the rectangular $$PQRS$$.

Then it passes through $$P\,\,(2,1)$$

$$\therefore$$ $${4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)$$

Also, given that, it passes through $$(4,0)$$

$$\therefore$$ $${{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16$$

$$ \Rightarrow {b^2} = 4/3$$ $$\left[ {\,\,} \right.$$ substituting $${{a^2} = 16\,\,}$$ in $$\left. {e{q^n}\left( a \right)\,\,} \right]$$

$$\therefore$$ The required ellipse is $${{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1$$

or $${x^2} + 12y{}^2 = 16$$

4

MCQ (Single Correct Answer)

A parabola has the origin as its focus and the line $$x=2$$ as the directrix. Then the vertex of the parabola is at

A

$$(0,2)$$

B

$$(1,0)$$

C

$$(0,1)$$

D

$$(2,0)$$

Vertex of a parabola is the mid point of focus and the point

where directrix meets the axis of the parabola.

Here focus is $$O\left( {0,0} \right)$$ and directrix meets the axis at $$B\left( {2,0} \right)$$

$$\therefore$$ Vertex of the parabola is $$(1,0)$$

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