1

### JEE Main 2016 (Online) 10th April Morning Slot

P and Q are two distinct points on the parabola, y2 = 4x, with parameters t and t1 respectively. If the normal at P passes through Q, then the minimum value of $t_1^2$ is :
A
2
B
4
C
6
D
8

## Explanation

t1 = $-$ t $-$ ${2 \over t}$

$t_1^2$ = t2 + ${4 \over {{t^2}}}$ + 4

t2 + ${4 \over {{t^2}}}$ $\ge$ 2$\sqrt {{t^2}.{4 \over {{t^2}}}} = 4$

Minimum value of $t_1^2$ = 8
2

### JEE Main 2016 (Online) 10th April Morning Slot

A hyperbola whose transverse axis is along the major axis of the conic, ${{{x^2}} \over 3} + {{{y^2}} \over 4} = 4$ and has vertices at the foci of this conic. If the eccentricity of the hyperbola is ${3 \over 2},$ then which of the following points does NOT lie on it ?
A
(0, 2)
B
$\left( {\sqrt 5 ,2\sqrt 2 } \right)$
C
$\left( {\sqrt {10} ,2\sqrt 3 } \right)$
D
$\left( {5,2\sqrt 3 } \right)$

## Explanation

${{{x^2}} \over {12}} + {{{y^2}} \over {16}}$ = 1

e = $\sqrt {1 - {{12} \over {16}}}$ = ${1 \over 2}$

Foci (0, 2)   &   (0, $-$ 2)

So, transverse axis of hyperbola

= 2b = 4

$\Rightarrow$ b = 2 & a2 = 12 (e2 $-$ 1)

$\Rightarrow$   a2 = 4$\left( {{9 \over 4} - 1} \right)$

$\Rightarrow$   a2 = 5

$\therefore$    It's equation is ${{{x^2}} \over 5} - {{{y^2}} \over 4}$ = $-$ 1

The point (5, 2$\sqrt 3$) does not satisfy the above equation.
3

### JEE Main 2017 (Offline)

A hyperbola passes through the point P$\left( {\sqrt 2 ,\sqrt 3 } \right)$ and has foci at $\left( { \pm 2,0} \right)$. Then the tangent to this hyperbola at P also passes through the point
A
$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$
B
$\left( {\sqrt 3 ,\sqrt 2 } \right)$
C
$\left( { - \sqrt 2 , - \sqrt 3 } \right)$
D
$\left( {3\sqrt 2 ,2\sqrt 3 } \right)$

## Explanation

Equation of hyperbola is ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$

foci is (±2, 0)

$\Rightarrow$ ae = 2

$\Rightarrow$ a2e2 = 4

Since b2 = a2 (e2 – 1)

b2 = a2 e2 – a2

$\therefore$ a2 + b2 = 4 .....(1)

Also Hyperbola passes through $\left( {\sqrt 2 ,\sqrt 3 } \right)$

$\therefore$ ${2 \over {{a^2}}} - {3 \over {{b^2}}} = 1$

$\Rightarrow$ ${2 \over {4 - {b^2}}} - {3 \over {{b^2}}} = 1$

$\Rightarrow$ (b2 – 3) (b2 + 4) = 0

$\therefore$ b2 = 3 or b2 = -4

For b2 = 3

$\Rightarrow$ a2 = 1

${{{x^2}} \over 1} - {{{y^2}} \over 3} = 1$

Equation of tangent is ${{\sqrt 2 x} \over 1} - {{\sqrt 3 y} \over 3} = 1$

It satisfy point $\left( {2\sqrt 2 ,3\sqrt 3 } \right)$.
4

### JEE Main 2017 (Offline)

The eccentricity of an ellipse whose centre is at the origin is ${1 \over 2}$. If one of its directrices is x = – 4, then the equation of the normal to it at $\left( {1,{3 \over 2}} \right)$ is
A
2y – x = 2
B
4x – 2y = 1
C
4x + 2y = 7
D
x + 2y = 4

## Explanation

Given e = ${1 \over 2}$ and ${a \over e}$ = 4

$\therefore$ $a$ = 2

We have b2 = $a$2 (1 – e2) = $4\left( {1 - {1 \over 4}} \right)$ = 3

$\therefore$ Equation of ellipse is

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

Now, the equation of normal at $\left( {1,{3 \over 2}} \right)$ is

${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$

$\Rightarrow$ ${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$

$\Rightarrow$ 4x – 2y = 1