1

### 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 = ${{\sqrt 3 } \over 2}$
2

### 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
3

### JEE Main 2018 (Offline)

Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and $\angle$CPB = $\theta$, then a value of tan$\theta$ is :
A
${4 \over 3}$
B
${1 \over 2}$
C
2
D
3

## Explanation

As equation of tangent PA at (x1, y1) on the parabola y2 = 4ax,

yy1 = 2a (x + x1)

here (x1, y1) = ( 16, 16)

y . 16 = 2.4 (x + 16)

$\Rightarrow$ 2y = x + 16 .....(1)

At pont A value of y = 0

putting y = 0 in equation (1) we get,

0 = x + 16

$\Rightarrow$ x = $-$ 16

$\therefore\,\,\,$ Coordinate of point A = ($-$ 16, 0)

Slope of line P A :

As$\,\,\,\,$ 2y = x + 16

$\therefore\,\,\,$ y = ${1 \over 2}\,$ x + 8

$\therefore\,\,\,$ Slope (m) = ${1 \over 2}\,$

Let slope of perpendicular line PB passing through point p(16, 16) = m'

$\therefore\,\,\,$ m m' = $-$ 1

$\Rightarrow$ ${1 \over 2}$ $\times$ m' = $-$ 1

$\Rightarrow$ ' = $-$ 2

As Equation of normal PB, when slope is m,

y = mx $-$ 2am $-$am3

Here m = m' = $-$ 2 and a = 4

$\therefore\,\,\,$ y = $-$2x $-$ 2(4) ($-$2) $-$ 4 . ($-$2)3

$\Rightarrow$ y = $-$ 2x + 16 + 32

$\Rightarrow$ y = $-$ 2x + 48 ..... (2)

At point B, y = 0

puttig y = 0 at equation (2) we get,

0 = $-$ 2x + 48

$\Rightarrow$x = 24

$\therefore\,\,\,$ Coordinate of point B = (24, 0)

A circle is passing through point P, A and B, and C is the center of the circle.

So, AC and BC are the radius.

Then AC = BC, So C is the middle point of line AB.

$\therefore\,\,\,$ C = $\left( {{{24 - 16} \over 2},{{0 + 0} \over 2}} \right)$ = (4, 0)

$\angle$CPB = $\theta$ and we have to find tan $\theta$.

Slope of line PC = ${{16 - 0} \over {16 - 4}}$ = ${4 \over 3}$ =m1

and we know slope of line PB = $-$2 = m2

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

$\Rightarrow$ tan $\theta$ = $\left| {{{{4 \over 3} + 2} \over {1 - \left( {{4 \over 3}} \right)\left( 2 \right)}}} \right|$

$\Rightarrow$ tan $\theta$ = $\left| {{{{{10} \over 3}} \over { - {5 \over 3}}}} \right|$

$\Rightarrow$ tan $\theta$ = $\left| { - 2} \right|$

$\Rightarrow$ tan $\theta$ = 2
4

### JEE Main 2018 (Online) 15th April Morning Slot

If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is :
A
x2 $-$ 4y2 + 16x2y2 = 0
B
x2 $-$ 4y2 $-$ 16x2y2 = 0
C
4x2 $-$ y2 + 16x2y2 = 0
D
4x2 $-$ y2 $-$ 16x2y2 = 0

## Explanation

Equation of hyperbola is :

4y2 = x2 + 1 $\Rightarrow$ $-$ x2 + 4y2 = 1

$\Rightarrow$ $\,\,\,$ $-$ ${{{x^2}} \over {{1^2}}}$ + ${{{y^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}}$ = 1

$\therefore$  a = 1, b = ${1 \over 2}$

Now, tangent to the curve at point (x1, y1) is given by

4 $\times$ 2y1${{dy} \over {dx}}$ = 2x1

$\Rightarrow$   ${{dy} \over {dx}}$ = ${{2{x_1}} \over {8{y_1}}}$ = ${{{x_1}} \over {4{y_1}}}$

Equation of tangent at (x1, y1) is

y = mx + c

$\Rightarrow$   y = ${{{x_1}} \over {4{y_1}}}$ . x + c

As tangent passes through (x1, y1)

$\therefore$   y1 = ${{{x_1}{x_1}} \over {4{y_1}}} + c$

$\Rightarrow$   C = ${{4y_1^2 - x_1^2} \over {4{y_1}}}$ = ${1 \over {4{y_1}}}$

Therefore, y = ${{{x_1}} \over {4{y_1}}}x + {1 \over {4{y_1}}}$

$\Rightarrow$ 4y1y = x1x + 1

which intersects x axis at A $\left( {{{ - 1} \over {{x_1}}},0} \right)$ and y axis at $B\left( {0,{1 \over {4{y_1}}}} \right)$

Let midpoint of AB is (h, k)

$\therefore$   h = ${{ - 1} \over {2{x_1}}}$

$\Rightarrow $$\,\,\, x1 = {{ - 1} \over {2h}} & y1 = {1 \over {8k}} Thus, 4{\left( {{1 \over {8k}}} \right)^2} = {\left( {{{ - 1} \over {2h}}} \right)^2} + 1 \Rightarrow$$\,\,\,$ ${1 \over {16{k^2}}}$ = ${1 \over {4{h^2}}}$ + 1

$\Rightarrow$ $\,\,\,$ 1 = ${{16{k^2}} \over {4{h^2}}}$ + 16k2

$\Rightarrow$$\,\,\,$ h2 = 4k2 + 16h2 k.

So, required equation is

x2 $-$ 4y2 $-$ 16x2 y2 = 0