1

### JEE Main 2017 (Online) 8th April Morning Slot

Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is ${3 \over 5}$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :
A
8
B
32
C
80
D
40

## Explanation

e = 3/5 & 2ae = 6  $\Rightarrow$   a = 5

$\because$   b2 = a2 (1 $-$ e2)

$\Rightarrow$   b2 = 25(1 $-$ 9/25)

$\Rightarrow$   b = 4

$\therefore$   area of required quadrilateral

= 4(1/2 ab) = 2ab = 40
2

### JEE Main 2017 (Online) 9th April Morning Slot

If y = mx + c is the normal at a point on the parabola y2 = 8x whose focal distance is 8 units, then $\left| c \right|$ is equal to :
A
$2\sqrt 3$
B
$8\sqrt 3$
C
$10\sqrt 3$
D
$16\sqrt 3$

## Explanation

c = $-$ 29m $-$ 9m3

a = 2

Given (at2 $-$ a)2 + 4a2t2 = 64

$\Rightarrow$   (a(t2 + 1)) = 8

$\Rightarrow$   t2 + 1 = 4 $\Rightarrow$ t2 = 3

$\Rightarrow$   t = $\sqrt 3$

$\therefore$   c = 2at(2 + t2)

= $2\sqrt 3 \left( 5 \right)$

$\left| c \right|$ = 10$\sqrt 3$
3

### JEE Main 2017 (Online) 9th April Morning Slot

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points (4, −1) and (−2, 2) is :
A
${1 \over 2}$
B
${2 \over {\sqrt 5 }}$
C
${{\sqrt 3 } \over 2}$
D
${{\sqrt 3 } \over 4}$

## Explanation

Centre at (0, 0)

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

at point (4, $-$ 1)

${{16} \over {{a^2}}} + {1 \over {{b^2}}}$ = 1

$\Rightarrow$   16b2 + a2 = a2b2         . . . .(i)

at point ($-$ 2, 2)

${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$

$\Rightarrow$   4b2 + 4a2 = a2b2          . . . .(ii)

$\Rightarrow$   16b2 + a2 = 4a2 + 4b2

From equations (i) and (ii)

$\Rightarrow$   3a2 = 12b2

$\Rightarrow$   a2 = 4b2

b2 = a2(1 $-$ e2)

$\Rightarrow$   e2 = ${3 \over 4}$

$\Rightarrow$   e = ${{\sqrt 3 } \over 2}$
4

### JEE Main 2018 (Offline)

Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q.

If these tangents intersect at the point T(0, 3) then the area (in sq. units) of $\Delta$PTQ is :
A
$36\sqrt 5$
B
$45\sqrt 5$
C
$54\sqrt 3$
D
$60\sqrt 3$

## Explanation

Here PQ is the chord of contact.

Equation of PQ is

x.0 $-$ y.3 = 36

$\Rightarrow \,\,\,\,y = - 12$

Putting value of y = $-$ 12 in the equation

4x2 $-$ y2 = 36

we get , 4x2 $-$ 144 = 36

$\Rightarrow \,\,\,\,4{x^2}\, = \,180$

$\Rightarrow \,\,\,\,\,{x^2}\,\, = \,\,45$

$\Rightarrow \,\,\,\,x = \pm \,3\sqrt 5$

$\therefore\,\,\,$ Coordinate of $P = \left( { - 3\sqrt 5 , - 12} \right)$ and

Coordinate of $Q = \left( {3\sqrt 5 , - 12} \right)$

$\therefore\,\,\,$ Length of PQ $= 3\sqrt 5 + 3\sqrt 5$

$= 6\sqrt 5 .$

TM is the height of the triangle Length of TM = 12 + 3 = 15

$\therefore\,\,\,$ Area of $\Delta PQT$ = ${1 \over 2} \times 6\sqrt 5 \times 15$

$= 45\sqrt 5$ sq. units