Question

The equation of the normal of slope m to the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
is given by $y=mx\pm \frac{m(a^2+b^2)}{\sqrt{a^2-b^2{m}^2}}$

• A. True
• B. False

Answer 1

The correct option is A

True

We already know that equation of normal at a point $(x_1,y_1)$ is given by,

$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

If m is the slope of normal, at $(x_1,y_1)$

then $m=-\frac{a^2{y}_1}{b^2{x}_1}$

since $(x_1,y_1)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

putting the value of $y_1$ in the equation

we get,

$x_1=\pm \frac{a^2}{\sqrt{a^2-b^2{m}^2}}$

$y_1=\pm \frac{b^{2}m}{\sqrt{a^2-b^2{m}^2}}$

$\therefore$ On giving these values in the equation

$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$ we get

$y=mx\pm \frac{m(a^2+b^2)}{\sqrt{a^2-b^2{m}^2}}$
as the equation of normal at the point
$(\pm \frac{a^2}{\sqrt{a^2-b^2{m}^2}},\pm \frac{b^{2}m}{\sqrt{a^2-b^2{m}^2}})$