1

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

A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :
A
${3 \over 2}$
B
$\sqrt 3$
C
2
D
${2 \over {\sqrt 3 }}$

## Explanation

Let the equation of hyperbola

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

Given 2a = 4

$\Rightarrow$  $a$ = 2

It passes through (4, 2)

$\therefore$  ${{16} \over 4} - {4 \over {{b^2}}}$ = 1

$\Rightarrow$  b2 = ${4 \over 3}$

e = $\sqrt {1 + {{{b^2}} \over {{a^2}}}}$ = $\sqrt {1 + {{4/3} \over 4}}$

= $\sqrt {1 + {1 \over 3}}$ = ${2 \over {\sqrt 3 }}$
2

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

Let A(4, $-$ 4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $\Delta$ACB is maximum. Then, the area (in sq. units) of $\Delta$ACB, is :
A
$31{1 \over 4}$
B
$30{1 \over 2}$
C
32
D
$31{3 \over 4}$

## Explanation

$\Delta ABC = {1 \over 2}\left| {\matrix{ 4 & { - 4} & 1 \cr 9 & 6 & 1 \cr {{t^2}} & {2t} & 1 \cr } } \right|$

D = 60 + 10t $-$ 10t2

${{d\Delta } \over {dt}} = 0 \Rightarrow t = {1 \over 2}$

${{{d^2}\Delta } \over {d{t^2}}} = - 20 < 0$

$\therefore$  max at $t = {1 \over 2}$

max area $\Delta = 65 - {5 \over 2}$

$= {{125} \over 2} = 31{1 \over 4}$
3

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

If the parabolas y2 = 4b(x – c) and y2 = 8ax have a common normal, then which on of the following is a valid choice for the ordered triad (a, b, c) ?
A
(1, 1, 3)
B
(1, 1, 0)
C
$\left( {{1 \over 2},2,0} \right)$
D
$\left( {{1 \over 2},2,3} \right)$

## Explanation

Normal to the two given curves are

y = m(x – c) – 2bm – bm3,

y = mx – 4am – 2am3

If they have a common normal, then

(c + 2b)m + bm3 = 4am + 2am3

$\Rightarrow$ (4a – c – 2b) m = (b – 2a)m3

$\Rightarrow$ (4a – c – 2b) = (b – 2a)m2

$\Rightarrow$ m2 = ${c \over {2a - b}} - 2$ $>$ 0

By checking all options we found (A) is right option.

Note : We get that all the options are correct for m = 0 i.e., when common normal is x-axis. But may be in question they want common normal other than x – axis, hence answer is (A).
4

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

The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is -
A
x $-$ y + 9 = 0
B
x $-$ y $-$ 3 = 0
C
x $-$ y + 1 = 0
D
x $-$ y + 7 = 0

## Explanation

Hyperbola ${{{x^2}} \over 5} - {{{y^2}} \over 4} = 1$

slope of tangent = 1

equation of tangent y = x $\pm$ $\sqrt {5 - 4}$

$\Rightarrow$  y = x $\pm$ 1

$\Rightarrow$  y = x + 1

or

$\Rightarrow$  y = x $-$ 1