A vertical line passing through the point (h,0) intersects the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ at the points P and Q. If the tangents to the ellipse at P and Q meet at the point R.
If ${Δ(h)= area of the ΔPQR,Δ}_{1} = \underset{1 / 2 \leq h \leq 1}{m a x} Δ (h)$ ${and Δ}_{2} = min_{1 / 2 \leq h \leq 1} Δ (h) t h e n \frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} =$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 9.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Δ_{1} = Δ_{max} occurs at \cos θ = \frac{1}{4} = (\frac{2 \sqrt{3} \sin^{3} θ}{\cos θ})$
When $\cos θ = \frac{1}{4} = \frac{45 \sqrt{5}}{8} Δ_{2} = Δ_{min}$ occurs at $\cos θ = \frac{1}{2} = (\frac{2 \sqrt{3} \sin^{3} θ}{\cos θ})$
When $\cos θ = \frac{1}{2} = \frac{9}{2} ∴ \frac{8}{\sqrt{5}} Δ_{1} - Δ_{2} = 45 - 36 = 9$