JEE Advanced · Mathematics · 14. Ellipse

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 $Q$ . Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$ . If $Δ (h) =$ area of triangle $PQR, Δ_{1} = \max_{\frac{1}{2} \leq h \leq 1} Δ (h)$ and $Δ_{2} = \min_{\frac{1}{2} \leq h \leq 1} Δ (h)$ , then $\frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} =$

A 3
B 6
C 9
D 12

Verified Solution

Answer & Solution

Correct Answer

(C) 9

Step-by-step Solution

Detailed explanation

Point of intersection of tangents at P and Q is

R (2 sec θ, 0)

Area of

Δ PQR = \frac{1}{2} \cdot 2 \sqrt{3} sin θ \cdot (2 sec θ - 2 cos θ)

\Rightarrow Δ = 2 \sqrt{3} \cdot \frac{{sin}^{3} θ}{cos θ}

; where

cos θ \in [\frac{1}{4}, \frac{1}{2}]

Now

\frac{d Δ}{d θ} = \frac{2 \sqrt{3} |cos θ \cdot {3 sin}^{2} θ cos θ - {sin}^{3} θ (- sin θ)|}{{cos}^{2} θ} > 0

θ

increases,

Δ

increases

\Rightarrow

when cos

θ

decreases,

Δ

increases

∴ Δ_{min} occurs at cos θ = 1 / 2, Therefore Δ_{2} = 2 \sqrt{3} \cdot \frac{{(1 - 1 / 4)}^{3 / 2}}{1 / 2} = 4 \sqrt{3} \cdot \frac{3 \sqrt{3}}{8} = \frac{36}{8}

Δ_{max} occurs at cos θ = 1 / 4, Therefore Δ_{1} = 2 \sqrt{3} \cdot \frac{{(1 - 1 / 16)}^{3 / 2}}{1 / 4} = 8 \sqrt{3} \cdot \frac{15 . \sqrt{15}}{4.4.4} = \frac{2 \sqrt{3} . 15 . \sqrt{3} \sqrt{5}}{16}

\Rightarrow Δ_{1} = \frac{45}{8} \sqrt{5}

Now

\frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} = 45 - 36 = 9

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A vertical line passing through the point h , 0 intersects the ellipse x 2 4 + y 2 3 = 1 at the points P and Q . Let the tangents to the ellipse at P and Q meet at the point R . If Δ(h)= area of triangle PQR,Δ1=max12≤h≤1Δh and Δ2=min12≤h≤1Δ(h), then 85Δ1-8Δ2=

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