If the curves frac{x^2}{a^2}+frac{y^2}{b^2}=1 | vedclass

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(A) We know that two conics $\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1$ and $\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=1$ intersect orthogonally if and only if

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$9$

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(A) We know that two conics $\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1$ and $\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=1$ intersect orthogonally if and only if $a_1^2-b_1^2 = a_2^2-b_2^2$, which can be rearranged as $a_1^2-a_2^2 = b_1^2-b_2^2$. Given the curves are $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$. Applying the condition for orthogonal intersection: $a^2-25 = b^2-16$ Rearranging the terms to find $a^2-b^2$: $a^2-b^2 = 25-16$ $a^2-b^2 = 9$ Thus, the value is $9$.

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

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