Q64 / 160 · AP EAMCET 2014 · Mathematics
If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut each other orthogonally, then $a^2-b^2$ equals to
- Correct answerA$9$
- B$400$
- C$75$
- D$41$
Solution
We know, two curves $\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1$ and $\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=1$ cut orthogonally.
Then, $a_1^2-a_2^2=b_1^2-b_2^2$
Here, equation of curves are
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { and } \frac{x^2}{25}+\frac{y^2}{16}=1
$$
$\therefore$ Condition of orthogonally,
$$
\begin{aligned}
\Rightarrow \quad & a^2-25=b^2-16 \\
\Rightarrow \quad & a^2-b^2=25-16 \\
& a^2-b^2=9
\end{aligned}
$$
Hence, the correct answer is option: 'A'.