JEE Main 2018 — Parabola Question with Solution

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 = $$\frac{1}{2}$$ x + 8

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

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

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

$$\Rightarrow$$ $$\frac{1}{2}$$ $$\times$$ 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 = $$\left(\frac{24-16}{2},\frac{0+0}{2}\right)$$ = (4, 0)

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

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

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

$$\therefore$$ tan $$\theta$$ = $$\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$$

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

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

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

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