Search any question & find its solution
Question:
Answered & Verified by Expert
The equation of the hyperbola, whose eccentricity is $\sqrt{2}$ and whose foci are 16 units apart, is
Options:
Solution:
1057 Upvotes
Verified Answer
The correct answer is:
$x^2-y^2=32$
Let the equation of hyperbola be
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
We have,
$e=\sqrt{2} \text { and } 2 a e=16 \Rightarrow 2 a(\sqrt{2})=16$
$\Rightarrow \quad a=\frac{16}{2 \sqrt{2}}=4 \sqrt{2}$
$\therefore$ Equation of hyperbola is
$\frac{x^2}{32}-\frac{y^2}{32}=1 \Rightarrow x^2-y^2=32$
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
We have,
$e=\sqrt{2} \text { and } 2 a e=16 \Rightarrow 2 a(\sqrt{2})=16$
$\Rightarrow \quad a=\frac{16}{2 \sqrt{2}}=4 \sqrt{2}$
$\therefore$ Equation of hyperbola is
$\frac{x^2}{32}-\frac{y^2}{32}=1 \Rightarrow x^2-y^2=32$
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.