Question

The equation to the normal to the hyperbola $\frac {x^2}{16}- \frac {y^2}{9}=1$ at $(-4,0)$ is

• A. 2x - 3y = 1
• B. x = 0
• C. x = 1
• D. y = 0

Answer 1

Answer: (D)

y = 0

Answer 2

We know that, the equation of normal at the point $\left(x_{1}, y_{1}\right)$ to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2}$ Given equation is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ Here, $\,\,\,\,\,\,a^{2}=16,\,\, b^{2}=9$ $\therefore\,\,\,$ The equation of normal at the point $(-4,0)$ is $\frac{16 x}{-4}+\frac{9 y}{0}=16+9$ $\Rightarrow\,\,\,\,\, \frac{9 y}{0}=25+\frac{16 x}{4}$ $\Rightarrow \,\,\,\,\,\,9 y=0$ $\Rightarrow \,\,\,\,y=0$