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 \(\Delta(h)=\) area of the triangle \(P Q R, \Delta_{1}=\max _{1 / 2 \leq h \leq 1} \Delta(h)\) and \(\Delta_{2}=\min _{1 / 2 \leq h \leq 1} \Delta(h)\), then \(\frac{8}{\sqrt{5}} \Delta_{1}-8 \Delta_{2}=\)

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · 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 the triangle

P QR, Δ_{1} = max_{1/2 \leq h \leq 1} Δ (h)

and

Δ_{2} = min_{1/2 \leq h \leq 1} Δ (h)

, then

\frac{8}{5} Δ_{1} - 8 Δ_{2} =

Answer: 9

The line

x = h

cuts the ellipse at

P = (h, y_{0})

and

Q = (h, - y_{0})

with

y_{0} = 3 (1 - h^{2} /4)

, and the tangents there meet on the major axis at

R = (4/ h, 0)

, the pole of the chord. The triangle

P QR

has base

P Q = 2 y_{0}

and height

\frac{4}{h} - h

, so its area is

Δ (h) = y_{0} (\frac{4}{h} - h) = \frac{3}{2 h} (4 - h^{2})^{3/2}

. On

\frac{1}{2} \leq h \leq 1

this is decreasing, so the maximum is at

h = \frac{1}{2}

, namely

Δ_{1} = \frac{45 5}{8}

, and the minimum is at

h = 1

, namely

Δ_{2} = \frac{9}{2}

. Then

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

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Integer answer