Find the condition for the line \(a x+b y+c=0\) to be a normal to an ellipse…
Find the condition for the line \(a x+b y+c=0\) to be a normal to an ellipse \(\frac{x^2}{4}+\frac{y^2}{36}=1\)
\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{144}{c^2}\)
\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{128}{c^2}\)
\(\frac{1}{a^2}+\frac{9}{b^2}=\frac{256}{c^2}\)
\(\frac{1}{a^2}+\frac{9}{b^2}=\frac{32}{c^2}\)
Solution
Let a point \(P(2 \cos \theta, 6 \sin \theta)\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{36}=1\), so equation of normal to the ellipse at point \(P\) is
\(\begin{aligned}
& \frac{x-2 \cos \theta}{\frac{\cos \theta}{2}}=\frac{y-6 \sin \theta}{\frac{\sin \theta}{6}} \\
& \Rightarrow \quad 2 x \sec \theta-4=6 y \operatorname{cosec} \theta-36 \\
& \Rightarrow \quad 2 x \sec \theta-6 y \operatorname{cosec} \theta+32=0 \quad \ldots (i)
\end{aligned}\)
Let normal (i) represent the line \(a x+b y+c=0\)
\(\begin{aligned}
& \text { So } \frac{a}{2 \sec \theta}=\frac{b}{-6 \operatorname{cosec} \theta}=\frac{c}{32} \\
& \Rightarrow \cos \theta=\frac{c}{16 a}, \sin \theta=-\frac{3 c}{16 b} \\
& \because \quad \cos ^2 \theta+\sin ^2 \theta=1 \\
& \Rightarrow \quad \frac{c^2}{256 a^2}+\frac{9 c^2}{256 b^2}=1 \Rightarrow \frac{1}{a^2}+\frac{9}{b^2}=\frac{256}{c^2}
\end{aligned}\)
Hence, option (c) is correct.