Search any question & find its solution
Question:
Answered & Verified by Expert
The products of lengths of perpendiculars from any point of hyperbola $x^2-y^2=8$ to its asymptotes, is
Options:
Solution:
1869 Upvotes
Verified Answer
The correct answer is:
$4$
We have the asymptotes are
$x^2-y^2=0 \Rightarrow x= \pm y$
i.e. $x+y=0$ and $x-y=0$
The product of length of perpendiculars
$=\left(\frac{x-y}{\sqrt{1^2+1^2}}\right)\left(\frac{x+y}{\sqrt{1^2+1^2}}\right)=\frac{x^2-y^2}{\sqrt{2} \cdot \sqrt{2}}=\frac{8}{2}=4$
$x^2-y^2=0 \Rightarrow x= \pm y$
i.e. $x+y=0$ and $x-y=0$
The product of length of perpendiculars
$=\left(\frac{x-y}{\sqrt{1^2+1^2}}\right)\left(\frac{x+y}{\sqrt{1^2+1^2}}\right)=\frac{x^2-y^2}{\sqrt{2} \cdot \sqrt{2}}=\frac{8}{2}=4$
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.