1

### JEE Main 2018 (Online) 16th April Morning Slot

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is ${3 \over 2}$ units, then its eccentricity is :
A
${1 \over 2}$
B
${1 \over 3}$
C
${2 \over 3}$
D
${1 \over 9}$

## Explanation

If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,

then distance between focus and vertex,

a $-$ ae = ${3 \over 2}$ (given)

$\Rightarrow$ $\,\,\,$ a (1 $-$ e) = ${3 \over 2}$

Length of latus rectum,

${{2{b^2}} \over a} = 4$

$\Rightarrow$ $\,\,\,$ b2 = 2a

$\Rightarrow$ $\,\,\,$ a2(1 $-$ e2) = 2a   [As b2 = a2 (1 $-$ e2)]

$\Rightarrow$ $\,\,\,$ a (1 $-$ e) ( 1 + e) = 2

Putting    a (1 $-$ e) = ${3 \over 2}$

$\Rightarrow$ $\,\,\,$ ${3 \over 2}$ (1 + e) = 2

$\Rightarrow$ $\,\,\,$ 3 + 3e = 4

$\Rightarrow$ $\,\,\,$ e = ${1 \over 3}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

Let P be a point on the parabola, x2 = 4y. If the distance of P from the center of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :
A
x + 4y $-$ 2 = 0
B
x $-$ y + 3 = 0
C
x + y +1 = 0
D
x + 2y = 0

## Explanation

Let P(2t, t2) be any point on the parabola.

Center of the given circle C = ($-$ g, $-$f) = ($-$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = ${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$ = ${{{t^2} - 0} \over {2t + 3}}$

Also, slope of tangent to parabola at P = ${{dy} \over {dx}}$ = ${x \over 2}$ = t

$\therefore$ Slope of normal = ${{ - 1} \over t}$

$\therefore$ ${{{t^2} - 0} \over {2t + 3}}$ = ${{ - 1} \over t}$

$\Rightarrow$ t3 + 2t + 3 = 0

$\Rightarrow$ (t+1) (t2 $-$ t + 3) = 0

$\therefore\,\,\,$ Real roots of above equation is

t = $-$ 1

Coordinate of P = (2t, t2) = ($-$2, 1)

Slope of tangent to parabola at P = t = $-$ 1

Therefore, equation of tangent is :

(y $-$ 1) = ($-$ 1) (x + 2)

$\Rightarrow$ x + y + 1 = 0
3

### JEE Main 2019 (Online) 9th January Morning Slot

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin, on the positive x-axis then which of the following points does not lie on it ?
A
(5, 2$\sqrt 6$)
B
(6, 4$\sqrt 2$)
C
(8, 6)
D
(4, -4)

## Explanation So the equation of the parabola,

${\left( {y - 0} \right)^2} = 4.a\left( {x - 2} \right)$

$\Rightarrow$  y2 = 4.2 (x $-$ 2)

$\Rightarrow$  y2 = 8 (x $-$ 2)

By checking each options you can see. point (8, 6) does not lie on the parabola.
4

### JEE Main 2019 (Online) 9th January Morning Slot

Let $0 < \theta < {\pi \over 2}$. If the eccentricity of the

hyperbola ${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$ = 1 is greater

than 2, then the length of its latus rectum lies in the interval :
A
(3, $\infty$)
B
$\left( {{3 \over 2},2} \right]$
C
$\left( {1,{3 \over 2}} \right]$
D
$\left( {2,3} \right]$

$\infty$

## Explanation

Given hyperbola,

${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$

here a = cos$\theta$

and b = sin$\theta$

We know, eccentricity of the hyperbola is,

$\sqrt {1 + {{{b^2}} \over {{a^2}}}}$

$\therefore$  Here eccentricity

(e) = $\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}}$

Given that,

$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$

$\Rightarrow$  $\sqrt {1 + {{\tan }^2}\theta } > 2$

$\Rightarrow$   1 + tan2$\theta$ > 4

$\Rightarrow$  tan2$\theta$ > 3

$\Rightarrow$  tan$\theta$ > $\pm \sqrt 3$

As given $\theta$ $\in$ $\left( {0,{\pi \over 2}} \right)$

possible value of tan$\theta$ > $\sqrt 3$

So, $\theta$ can be in the range ${\pi \over 3} < \theta < {\pi \over 2}$ We know latus ractum (LR) = ${{2{b^2}} \over a}$

$\therefore$  LR = ${{2{{\sin }^2}\theta } \over {\cos \theta }}$

= 2 tan$\theta$ sin$\theta$

We know in the range ${\pi \over 3} < \theta < {\pi \over 2}$ tan$\theta$ and sin$\theta$ both are increasing function.

So, at ${\pi \over 3}$ value of LR will be minimum and at ${\pi \over 2}$ value of LR will be maximum.

$\therefore$  Minimum value of LR = 2tan${\pi \over 3}$ sin${\pi \over 3}$

= 2 $\times$ $\sqrt 3 \times {{\sqrt 3 } \over 2}$
= 3

Maximum value of LR = 2tan${\pi \over 2}$ sin${\pi \over 2}$

= 2$\left( \infty \right) \times 1$

= $\infty$

$\therefore$  Internal of LR = (3, $\infty$)