JEE Main 2020 — Ellipse Question with Solution

Let foot of perpendicular is (h, k)

Given $$\frac{x^2}{4}+\frac{y^2}{2}=1$$

$$\therefore$$ a = 2, b = $$\sqrt{2}$$

and e = $$\sqrt{1-\frac{2}{4}}$$ = $$\frac{1}{\sqrt{2}}$$

$$\therefore$$ Focus(ae, 0) = $$\left(\sqrt{2},0\right)$$

Equation of tangent

y = mx + $$\sqrt{a^2{m}^2+b^2}$$

$$\Rightarrow$$ y = mx + $$\sqrt{4m^2+2}$$

Passes through (h, k)

(k – mh)² = 4m² + 2 .....(1)

Line perpendicular to tangent will have slope $$-\frac{1}{m}$$.

y - 0 = $$-\frac{1}{m}\left(x-\sqrt{2}\right)$$

my = -x + $$\sqrt{2}$$

$$\therefore$$ (h + mk)² = 2 .....(2)

Add equation (1) and (2)

k²(1 + m²) + h² (1 + m²) = 4 (1 + m²)

h² + k² = 4

x² + y² = 4 (Auxiliary circle)

$$\therefore$$ $$\left(-1,\sqrt{3}\right)$$ lies on the locus.