Question

Find the equation of normal to the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ at $(5,0)$.

• A. $y=0$
• B. $y=-1$
• C. $y=1$
• D. $y=-2$

Answer 1

The correct option is A $y=0$

We know equation of normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by, $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2-b^2$
Thus, required normal is, $5x+\frac{16y}{0}=9$
Clearly denominator of second term of L.H.S is $0$ so the equation of line is, $y=0$
Hence, option 'A' is correct.