Question

Paragraph :
Let $S$ and $S^{\prime }$ be the foci on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $P(x_1,y_1)$ be an arbitrary point as shown.
(b) Find the gradient of the tangent to the hyperbola at $P$.

• A. $\frac{a^2{x}_1}{b^2{y}_1}$
• B. $\frac{b^2{y}_1}{a^2{x}_1}$
• C. $\frac{b^2{x}_1}{a^2{y}_1}$
• D. $\frac{a^2{y}_1}{b^2{x}_1}$

Answer 1

The correct option is C $\frac{b^2{x}_1}{a^2{y}_1}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
i.e. $\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0$ (by implicit differentiation)
Therefore, $\frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$
Thus the gradient of the tangent at $P(x_1,y_1)$ is $\frac{b^2{x}_1}{a^2{y}_1}$.
Hence option C is correct.