Search any question & find its solution
Question:
Answered & Verified by Expert
If the equation $x+y+n=0$ represents a normal to the hyperbola $\frac{x^2}{6}-\frac{y^2}{2}=1$, then $n=$
Options:
Solution:
1990 Upvotes
Verified Answer
The correct answer is:
$\pm {4}$
$$
\begin{aligned}
& \text {} x+y+n=0 \\
& \Rightarrow y=-x-n \\
& m=-1, c=-n
\end{aligned}
$$
If $y=m x+c$ is normal then
$\begin{aligned} & c= \pm \frac{\left(a^2+b^2\right) m}{\sqrt{a^2-b^2 m^2}} \\ \therefore \quad & n^2=\frac{(6+2)^2(-1)^2}{\left(\sqrt{6-2(-1)^2}\right)^2} \\ \Rightarrow & n^2=\frac{64}{4}=16 \\ \therefore \quad & n= \pm 4 .\end{aligned}$
\begin{aligned}
& \text {} x+y+n=0 \\
& \Rightarrow y=-x-n \\
& m=-1, c=-n
\end{aligned}
$$
If $y=m x+c$ is normal then
$\begin{aligned} & c= \pm \frac{\left(a^2+b^2\right) m}{\sqrt{a^2-b^2 m^2}} \\ \therefore \quad & n^2=\frac{(6+2)^2(-1)^2}{\left(\sqrt{6-2(-1)^2}\right)^2} \\ \Rightarrow & n^2=\frac{64}{4}=16 \\ \therefore \quad & n= \pm 4 .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.