Question

PQ and QR are two focal chords of an ellipse and the

eccentric angles of P,Q,R and 2a, 2b, 2g respectively then tan

b tan g is equal to –

• A. cota
• B. cot²a
• C. 2 cot a
• D. None of these

Answer 1

Answer: (B)

cot²a

Answer 2

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

P (a cos 2a, b sin 2a), Q (a cos 2b , b sin 2 b)

R (a cos 2g, b sin 2g)

chord's PQ equation

$\frac{x}{a}$cos (a + b) + $\frac{y}{b}$sin (a + b) = cos (a – b)

PQ passes through the focus (ae, 0)

e = $\frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)}$

PR passes through the focus (– ae, 0) the

– e = $\frac{\cos(\alpha - \gamma)}{\cos(\alpha + \gamma)}$

$\frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)}$ = – $\frac{\cos(\alpha - \gamma)}{\cos(\alpha + \gamma)}$

Apply componendo and dividendo, we get

$\frac{\cos(\alpha + \beta) + \cos(\alpha - \beta)}{\cos(\alpha + \beta) - \cos(\alpha - \beta)}$ = $\frac{\cos(\alpha + \gamma) - \cos(\alpha - \gamma)}{\cos(\alpha + \gamma) + \cos(\alpha - \gamma)}$

$\frac{2\cos\alpha\cos\beta}{2\sin\alpha\sin\beta}$ = $\frac{2\sin\alpha\sin\gamma}{2\cos\alpha\cos\gamma}$

tan b tan g = cot² a