Question

e and $e^1$ are the eccentricities of the hyperbolas $16x^2-9y^2=144$ and $9x^2-16y^2$= - 144 then e - $e^1$ =

• A. 2
• B. 1
• C. 0
• D. $\frac{3}{2}$

Answer 1

The correct option is C

0

$\begin{array}{cc}16x^2-9y^2=144& 9x^2-16y^2=-144\\ \frac{x^2}{9}-\frac{y^2}{16}=1& \frac{x^2}{16}-\frac{y^2}{9}=-1\\ e=\sqrt{1+\frac{16}{9}}& e^1=\sqrt{1+\frac{16}{9}}\\ =\frac{5}{3}& \frac{5}{3}\end{array}$

= e = $e^1$ = 0