 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2009

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

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

### AIEEE 2008

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
$$(2,0)$$

## Explanation 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 $$(1,0)$$
3

### AIEEE 2008

A focus of an ellipse is at the origin. The directrix is the line $$x=4$$ and the eccentricity is $${{1 \over 2}}$$. Then the length of the semi-major axis is
A
$${{8 \over 3}}$$
B
$${{2 \over 3}}$$
C
$${{4 \over 3}}$$
D
$${{5 \over 3}}$$

## Explanation Perpendicular distance of directrix from focus

$$= {a \over e} - ae = 4$$

$$\Rightarrow a\left( {2 - {1 \over 2}} \right) = 4$$

$$\Rightarrow a = {8 \over 3}$$

$$\therefore$$ Semi major axis $$=8/3$$
4

### AIEEE 2007

The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a
A
circle
B
hyperbola
C
ellipse
D
parabola

## Explanation

Equation of normal at $$P\left( {x,y} \right)$$ is $$Y - y = - {{dx} \over {dy}}\left( {x - x} \right)$$

Coordinate of $$G$$ at $$X$$ axis is $$\left( {X,0} \right)$$ (let)

$$\therefore$$ $$0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x$$

$$\Rightarrow X = x + y{{dy} \over {dx}}$$ $$\therefore$$ Co-ordinate of $$G\left( {x + y{{dy} \over {dx}},0} \right)$$

Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$

as distance cannot be $$-ve,$$ therefore abscissa $$x$$ should be $$+ve$$

$$\therefore$$ $$x + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx$$

On Integrating $$\Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}$$

$$\therefore$$ the curve is a hyperbola

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