KCET · Maths · Hyperbola
The equation of the normal to the hyperbola \(\frac{x^{2}}{16}-
\frac{y^{2}}{9}=1\) at \((-4,0)\) is
- A \(2 \mathrm{x}-3 \mathrm{y}=1\)
- B \(\mathrm{x}=0\)
- C \(x=1\)
- D \(y=0\)
Answer & Solution
Correct Answer
(D) \(y=0\)
Step-by-step Solution
Detailed explanation
We know that, the equation of normal at the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) to the hyperbola \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) 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, \(\quad a^{2}=16, b^{2}=9\)
\(\therefore\) The equation of normal at the point \((-4,0)\) is
\[
\begin{aligned}
\frac{16 x}{-4}+\frac{9 y}{0} &=16+9 \\
\Rightarrow \quad \frac{9 y}{0} &=25+\frac{16 x}{4} \\
9 y &=0 \Rightarrow y=0
\end{aligned}
\]
\[
\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2}
\]
Given equation is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)
Here, \(\quad a^{2}=16, b^{2}=9\)
\(\therefore\) The equation of normal at the point \((-4,0)\) is
\[
\begin{aligned}
\frac{16 x}{-4}+\frac{9 y}{0} &=16+9 \\
\Rightarrow \quad \frac{9 y}{0} &=25+\frac{16 x}{4} \\
9 y &=0 \Rightarrow y=0
\end{aligned}
\]
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