Question

The equation of normal to a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at a point with parameter $\theta$ is

$axcos\theta -bycot\theta =a^2+b^2$

• A. True
• B. False

Answer 1

Answer: (B)

The given equation of hyperbola is,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ............(1)

The slope of the curve at a point is given by $\frac{dy}{dx}$.

Differentiating (1) with respect to x

$\frac{2x}{a^2}-\frac{2y}{b^2}.\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{x{b}^2}{y{a}^2}$

$\therefore$ slope of normal =$-\frac{y{a}^2}{x{b}^2}$

Slope of normal at $(asec\theta ,btan\theta ))$

$=-\frac{btan\theta .a^2}{asec\theta .b^2}$

$=-\frac{a}{b}sin\theta$

$\therefore$ Equation of normal is,

$y-btan\theta =-\frac{a}{b}.sin\theta (x-asec\theta )$

$by-b^{2}tan\theta =-ax.sin\theta +a^2.tan\theta$

$by+axsin\theta =tan\theta (a^2+b^2)$

$ax.cos\theta +bycot\theta =a^2+b^2$

Hence given equation is incorrect.