JEE Main 2019 — Parabola Question with Solution

For this parabola y² = 16x,

a = 4

Here PQ is focal cord.

Let P(at₁², 2at₁) and Q(at₂², 2at₂).

Given P(1, 4),

$$\therefore$$ at₁² = 1

$$\Rightarrow$$ 4t₁² = 1

$$\Rightarrow$$ t₁² = $$\frac{1}{4}$$

$$\Rightarrow$$ t₁ = $$\frac{1}{2}$$

In parabola if the parameter of one end point of the focal cord is t₁ then parameter of the other end point t₂ = $$-\frac{1}{t_1}$$

Here parameter for point Q t₂ = - 2

$$\therefore$$ Length of focal cord

|PQ| = a$${\left(t_1-t_2\right)}^2$$ = 4$${\left(\frac{1}{2}+2\right)}^2$$ = 25