Question

If the two tangents drawn on hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$ = 1 in such a way that the product of their gradients is c², then they intersects on the curve

• A. y² + b² = c² (x² – a²)
• B. y² + b² = c² (x² + a²)
• C. ax² + by² = c²
• D. None of these

Answer 1

Answer: (A)

y² + b² = c² (x² – a²)

Answer 2

Let (h,k) be the point of intersection. By SS₁ = T²,

$\left( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} - 1 \right)\left( \frac{h^{2}}{a^{2}} - \frac{k^{2}}{b^{2}} - 1 \right) = \left\lbrack \frac{hx}{a^{2}} - \frac{ky}{b^{2}} - 1 \right\rbrack^{2}$

⇒ x²

$\left\lbrack \frac{h^{2}}{a^{4}} - \frac{k^{2}}{a^{2}b^{2}} - \frac{1}{a^{2}} - \frac{h^{2}}{a^{4}} \right\rbrack - y^{2}\left\lbrack \frac{h^{2}}{a^{2}b^{2}} - \frac{k^{2}}{b^{4}} - \frac{1}{b^{2}} + \frac{k^{2}}{b^{4}} \right\rbrack$+ ..... = 0

We know that m₁ m₂ = $\frac{Coefficientofx^{2}}{Coefficientofy^{2}}$

⇒ m₁m₂ = $\frac{\frac{k^{1}}{a^{2}b^{2}} + \frac{1}{a^{2}}}{\frac{h^{2}}{a^{2}b^{2}} - \frac{1}{b^{2}}}$ = c²

⇒ $\left( \frac{k^{2} + b^{2}}{h^{2} - a^{2}} \right)$= c² or (y² + b² = c² (x² – a²)