Question 1511
CBSE Class 11 Mathematics · Conic Sections · Hyperbolas
For the hyperbola \(16x^2-9y^2=576\Rightarrow\frac{x^2}{36}-\frac{y^2}{64}=1\), determine its foci, vertices, eccentricity, and latus-rectum length.
Solution to Question 1511
Complete workingTranslate every condition into an equation and retain exact values throughout the hyperbola calculation.
\[\begin{aligned}16x^2-9y^2 &= 576\Rightarrow\frac{x^2}{36}-\frac{y^2}{64}\\&= 1\end{aligned}\]
\[\begin{aligned}a &= 6,b\\&= 8,c\\&= 10\end{aligned}\]
\[\begin{aligned}c^2 &= a^2+b^2,\quad e\\&= \frac ca,\quad\ell\\&= \frac{2b^2}{a}\end{aligned}\]
\[\begin{aligned}F &= (\pm10,0),\ V\\&= (\pm6,0),\ e\\&= \frac53,\ \ell\\&= \frac{64}{3}\end{aligned}\]
The last line satisfies the original conditions and supplies every requested part.
Final answer
\[F=(\pm10,0),\ V=(\pm6,0),\ e=\frac53,\ \ell=\frac{64}{3}\]