1

### JEE Main 2019 (Online) 10th January Evening Slot

Let S = $\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$ Then S represents
A
an ellipse whose eccentricity is ${1 \over {\sqrt {r + 1} }},$ where r > 1
B
an ellipse whose eccentricity is ${2 \over {\sqrt {r + 1} }},$ where 0 < r < 1
C
an ellipse whose eccentricity is ${2 \over {\sqrt {r - 1} }},$ where 0 < r < 1
D
an ellipse whose eccentricity is ${2 \over {\sqrt {r + 1} }},$ where r > 1

## Explanation

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

for r > 1,     ${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$

$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)}$

$= \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}}$

$= \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}}$
2

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

Equation of a common tangent to the parabola y2 = 4x and the hyperbola xy = 2 is
A
x + y + 1 = 0
B
4x + 2y + 1 = 0
C
x – 2y + 4 = 0
D
x + 2y + 4 = 0

## Explanation

Let the equation of tangent to parabola

y2 = 4x be y = mx + ${1 \over m}$

It is also a tangent to hyperbola xy = 2

$\Rightarrow$  x$\left( {mx + {1 \over m}} \right)$ = 2

$\Rightarrow$  x2m + ${x \over m}$ $-$ 2 = 0

D = 0 $\Rightarrow$   m = $-$ ${1 \over 2}$

So tangent is 2y + x + 4 = 0
3

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

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

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

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

### 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 ${{{x^2}} \over {64}} + {{{y^2}} \over {32}} = 1$