1

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

If $\theta$ denotes the acute angle between the curves, y = 10 – x2 and y = 2 + x2 at a point of their intersection, the |tan $\theta$| is equal to :
A
$8 \over 15$
B
$4 \over 9$
C
$7 \over 17$
D
$8 \over 17$

## Explanation Angle between the curves is the acute angle between the tangents at the point of intersection.

y = 10 $-$ x2 (for curve 1)

and y = 2 + x2 (for curve 2)

$\therefore$  10 $-$ x2 = 2 + x2

$\Rightarrow$  2x2 = 8

$\Rightarrow$  x2 = 4

$\Rightarrow$  x = 2, $-$ 2

$\therefore$  points of intersection (2, 6) and ($-$ 2, 6)

${{dy} \over {dx}}$ for curve 1 = $-$ 2x

$\therefore$  Slope(m1) of curve 1 is = $-$ 2(2) = $-$ 4

${{dy} \over {dx}}$ for curve 2 = 2x

$\therefore$  slope (m2) of curve 2 = 2 $\times$ 2 = 4

$\therefore$  tan$\theta$ = $\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$

= $\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|$

= ${8 \over {15}}$
2

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

Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x is :
A
$2\sqrt 3$y = 12x + 1
B
$\sqrt 3$y = x + 3
C
$2\sqrt 3$y = -x - 12
D
$\sqrt 3$y = 3x + 1

## Explanation

We know,

Equation of tangent to the parabola y2 = 4ax is,

y = mx + ${a \over m}$

$\therefore$  Equation of tangent to the parabola y2 = 4x is,

y = mx + ${1 \over m}$

$\Rightarrow$  m2x $-$ ym + 1 = 0

This tangent is also the tangent to the circle x2 + y2 $-$ 6x = 0

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$\therefore$  $\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$

$\Rightarrow$  (3m2 + 1)2 = 9(m4 + m2)

$\Rightarrow$  9m4 + 6m2 + 1 = 9m4 + 9m2

$\Rightarrow$  3m2 = 1

$\Rightarrow$  m = $\pm$ ${1 \over {\sqrt 3 }}$

So, possible tangents are

y = ${1 \over {\sqrt 3 }}$x + $\sqrt 3$

$\Rightarrow$  $\sqrt 3$y = x + 3

or   y = $-$ ${x \over {\sqrt 3 }}$ $-$ $\sqrt 3$

$\Rightarrow$  $\sqrt 3 y$ = $-$ x $-$ 3
3

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

= $\sqrt {1 + {1 \over 3}}$ = ${2 \over {\sqrt 3 }}$
4

### 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 $\Delta = 65 - {5 \over 2}$

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