1

### JEE Main 2019 (Online) 11th January Morning Slot

If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :
A
${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$
B
${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$
C
${1 \over {4{x^2}}} + {1 \over {2{y^2}}} = 1$
D
${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$

## Explanation

Equation of general tangent on ellipse

${x \over {a\,\sec \theta }} + {y \over {b\cos ec\theta }} = 1$

$a = \sqrt 2 ,\,\,b = 1$

$\Rightarrow {x \over {\sqrt 2 \sec \theta }} + {y \over {\cos ec\theta }} = 1$

Let the midpoint be (h, k)

$h = {{\sqrt 2 \sec \theta } \over 2} \Rightarrow \cos \theta = {1 \over {\sqrt 2 h}}$

and $k = {{\cos ec\theta } \over 2} \Rightarrow \sin \theta = {1 \over {2k}}$

$\because$  ${\sin ^2}\theta + {\cos ^2}\theta = 1$

$\Rightarrow$  ${1 \over {2{h^2}}} + {1 \over {4{k^2}}} = 1$

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

### JEE Main 2019 (Online) 11th January Evening Slot

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it ?
A
$\left( {4\sqrt 2 ,2\sqrt 3 } \right)$
B
$\left( {4\sqrt 3 ,2\sqrt 3 } \right)$
C
$\left( {4\sqrt 3 ,2\sqrt 2 } \right)$
D
$\left( {4\sqrt 2 ,2\sqrt 2 } \right)$

## Explanation

${{2{b^2}} \over a} = 8$ and 2ae $=$ 2b

$\Rightarrow$  ${b \over a}$ = e and 1 $-$ e2 = e2 $\Rightarrow$ e $=$ ${1 \over {\sqrt 2 }}$

$\Rightarrow$  b = 4$\sqrt 2$ and a $=$ 8

So equation of ellipse is ${{{x^2}} \over {64}} + {{{y^2}} \over {32}} = 1$
3

### JEE Main 2019 (Online) 11th January Evening Slot

If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is :
A
${{13} \over 6}$
B
2
C
${{13} \over 12}$
D
${{13} \over 8}$

## Explanation

2b = 5 and 2ae = 13

b2 = a2(e2 $-$ 1) $\Rightarrow$  ${{25} \over 4}$ = ${{169} \over 4}$ $-$ a2

$\Rightarrow$  a $=$ 6 $\Rightarrow$  e $=$ ${{13} \over {12}}$
4

### JEE Main 2019 (Online) 11th January Evening Slot

If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is
A
$5\sqrt 5$
B
${\left( {10} \right)^{2/3}}$
C
$5\left( {{2^{1/3}}} \right)$
D
5

## Explanation

Vertex is (a2, 0)

y2 $=$ $-$(x $-$ a2) and x $=$ 0 $\Rightarrow$ (0, $\pm$ 2a)

Area of triangle is $= {1 \over 2}.$4a.(a2) = 250

$\Rightarrow$  a3 = 125 or a = 5