 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

The eccentricity of an ellipse, with its centre at the origin, is $${1 \over 2}$$. If one of the directrices is $$x=4$$, then the equation of the ellipse is:
A
$$4{x^2} + 3{y^2} = 1$$
B
$$3{x^2} + 4{y^2} = 12$$
C
$$4{x^2} + 3{y^2} = 12$$
D
$$3{x^2} + 4{y^2} = 1$$

## Explanation

$$e = {1 \over 2}.\,\,$$ Directrix, $$x = {a \over e} = 4$$

$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$

$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3$$

Equation of elhipe is

$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$
2

### AIEEE 2004

If $$a \ne 0$$ and the line $$2bx+3cy+4d=0$$ passes through the points of intersection of the parabolas $${y^2} = 4ax$$ and $${x^2} = 4ay$$, then
A
$${d^2} + {\left( {3b - 2c} \right)^2} = 0$$
B
$${d^2} + {\left( {3b + 2c} \right)^2} = 0$$
C
$${d^2} + {\left( {2b - 3c} \right)^2} = 0$$
D
$${d^2} + {\left( {2b + 3c} \right)^2} = 0$$

## Explanation

Solving equations of parabolas

$${y^2} = 4ax$$ and $${x^2} = 4ay$$

we get $$(0,0)$$ and $$(4a, 4a)$$

Substituting in the given equation of line

$$2bx+3cy+4d=0,$$

we get $$d=0$$

and $$2b+3c=0$$ $$\Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$
3

### AIEEE 2003

The foci of yhe ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the hyperbola $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$ coincide. Then the value of $${b^2}$$ is
A
$$9$$
B
$$1$$
C
$$5$$
D
$$7$$

## Explanation

$${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$

$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$

$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$

$$\therefore$$ Foci $$= \left( { \pm 3,0} \right)$$

$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola

$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$

$$\therefore$$ $$e = {3 \over 4}$$

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

$$\Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$
4

### AIEEE 2003

The normal at the point$$\left( {bt_1^2,2b{t_1}} \right)$$ on a parabola meets the parabola again in the point $$\left( {bt_2^2,2b{t_2}} \right)$$, then
A
$${t_2} = {t_1} + {2 \over {{t_1}}}$$
B
$${t_2} = -{t_1} - {2 \over {{t_1}}}$$
C
$${t_2} = -{t_1} + {2 \over {{t_1}}}$$
D
$${t_2} = {t_1} - {2 \over {{t_1}}}$$

## Explanation

Equation of the normal to a parabola $${y^2} = 4bx$$ at point

$$\left( {bt_1^2,2b{t_1}} \right)$$ is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$

As given, it also passes through $$\left( {bt_2^2,2b{t_2}} \right)$$ then

$$2b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3$$

$$2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)$$

$$= - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)$$

$$\Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)$$

$$\Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}$$

$$\Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}$$

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