JEE Main 2019 — Ellipse Question with Solution

Given, Equation of ellipse 4x² + y² = 8

We know equation of tangent at any point (x₁, y₁) is

4xx₁ + yy₁ = 8

$$\therefore$$ Equation of tangent at point (1, 2) is

4x + 2y = 8

$$\Rightarrow$$ 2x + y = 4

$$\therefore$$ Slope of this tangent = -2

Equation of tangent at point (a, b) is

4ax + by = 8

Slope of this tangent = $$-\frac{4a}{b}$$

As tangent at (1, 2) and (a, b) are perpendicular

$$\therefore$$ $$\left(-2\right)\left(-\frac{4a}{b}\right)=-1$$

$$\Rightarrow$$ b = -8$$a$$

As point (a, b) lies on the ellipse

$$\therefore$$ $$4a^2+b^2=8$$

$$\Rightarrow$$ $$4a^2+64a^2=8$$

$$\Rightarrow$$ $$a^2=\frac{8}{68}$$ = $$\frac{2}{17}$$