JEE Main 2018 — Hyperbola Question with Solution

Given, 4x² $$-$$ 9y² = 36

After differentiating w.r.t.x, we get

4.2x $$-$$ 9.2.y.$$\frac{dy}{dx}$$ = 0

$$\Rightarrow$$ Slope of tangent = $$\frac{dy}{dx}$$ = $$\frac{4x}{9y}$$

So, slope of normal = $$\frac{-9y}{4x}$$

Now, equation of normal at point (x₀, y₀) is given by

y $$-$$ y₀ = $$\frac{-9y_0}{4x_0}$$ (x $$-$$ x₀)

As normal intersects X axis at A, Then

A = $$\left(\frac{13x_0}{9},0\right)$$ and B $$\equiv$$ $$\left(0,\frac{13y_0}{4}\right)$$

As OABP is parallelogram

$$\therefore$$ midpoint of OB $$\equiv$$ $$\left(0,\frac{13y_0}{8}\right)$$ $$\equiv$$ Midpoint of AP

So, P(x, y) $$\equiv$$ $$\left(\frac{-13x_0}{9},\frac{13y_0}{4}\right)$$     ...(i)

$$\because$$ (x₀, y₀) lies on hyperbola, therefore

4(x₀)² $$-$$ 9(y₀)² = 36

From equation (i) : x₀ = $$\frac{-9x}{13}$$ and y₀ = $$\frac{4y}{13}$$

From equation (ii), we get

9x² $$-$$ 4y² = 169

Hence, locus of point P is : 9x² $$-$$ 4y² = 169