If the curves x2a2+y2b2=1 and x225+y216=1 cut each other orthogonally, then a2−b2=

x2a2+y2b2=x225+y216⇒x2(1a2−12)=y2(116−1b2)x2a2+y2b2=1⇒2xa2+2yy1b2=0⇒yy1b2=−xa2⇒y1=−b2xa2y⇒m1=−b2xa2y Two curves cut each other orthogonally ⇒m1m2=−1⇒(−b2xa2y)(−16x25y)=−1⇒16b2x225a2y2=−1⇒16b225a2(1/16−1/221/a2−1/25)=−1 ⇒b2−16=−25+a2⇒a2−b2=9

If the curves x2a2+y2b2=1 and x225+y216=1 cut each other orthogonally, then a2−b2=

x2a2+y2b2=x225+y216⇒x2(1a2−12)=y2(116−1b2)x2a2+y2b2=1⇒2xa2+2yy1b2=0⇒yy1b2=−xa2⇒y1=−b2xa2y⇒m1=−b2xa2y Two curves cut each other orthogonally ⇒m1m2=−1⇒(−b2xa2y)(−16x25y)=−1⇒16b2x225a2y2=−1⇒16b225a2(1/16−1/221/a2−1/25)=−1 ⇒b2−16=−25+a2⇒a2−b2=9

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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 $a^{2} - b^{2} =$

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$400$

$75$

$41$

$9$

answer is D.

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Detailed Solution

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{x^{2}}{25} + \frac{y^{2}}{16} \Rightarrow x^{2} (\frac{1}{a^{2}} - \frac{1}{2}) = y^{2} (\frac{1}{16} - \frac{1}{b^{2}})$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{2 x}{a^{2}} + \frac{2 y y_{1}}{b^{2}} = 0 \Rightarrow \frac{y y_{1}}{b_{2}} = \frac{- x}{a^{2}} \Rightarrow y_{1} = \frac{- b^{2} x}{a^{2} y} \Rightarrow m_{1} = \frac{- b^{2} x}{a^{2} y}$
Two curves cut each other orthogonally $\Rightarrow m_{1} m_{2} = - 1$
$\Rightarrow (\frac{- b^{2} x}{a^{2} y}) (\frac{- 16 x}{25 y}) = - 1 \Rightarrow \frac{16 b^{2} x^{2}}{25 a^{2} y^{2}} = - 1 \Rightarrow \frac{16 b^{2}}{25 a^{2}} (\frac{1 / 16 - 1 / 22}{1 / a^{2} - 1 / 25}) = - 1$
$\Rightarrow b^{2} - 16 = - 25 + a^{2} \Rightarrow a^{2} - b^{2} = 9$