JEE Main 2018 — Hyperbola Question with Solution

Equation of hyperbola is :

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

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

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

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

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

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

Equation of tangent at (x₁, y₁) is

y = mx + c

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

As tangent passes through (x₁, y₁)

$$\therefore$$   y₁ = $$\frac{x_1{x}_1}{4y_1}+c$$

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

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

$$\Rightarrow$$ 4y₁y = x₁x + 1

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

Let midpoint of AB is (h, k)

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

$$\Rightarrow$$ x₁ = $$\frac{-1}{2h}$$ & y₁ = $$\frac{1}{8k}$$

Thus, 4$${\left(\frac{1}{8k}\right)}^2$$ = $${\left(\frac{-1}{2h}\right)}^2$$ + 1

$$\Rightarrow$$ $$\frac{1}{16k^2}$$ = $$\frac{1}{4h^2}$$ + 1

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

$$\Rightarrow$$ h² = 4k² + 16h² k.

So, required equation is

x² $$-$$ 4y² $$-$$ 16x² y² = 0