Question:
Let \(e_1\) and \(e_2\) be the eccentricities of the ellipse,
\(\frac{x^2}{25}+\frac{y^2}{b^2}=1\)
(\(b\lt5\)) and the hyperbola,
\(\frac{x^2}{16}-\frac{y^2}{b^2}=1\)
respectively satisfying \(e_1e_2=1\). If \(\alpha\) and \(\beta\) are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair \((\alpha,\beta)\) is equal to :
Let \(e_1\) and \(e_2\) be the eccentricities of the ellipse,
\(\frac{x^2}{25}+\frac{y^2}{b^2}=1\)
(\(b\lt5\)) and the hyperbola,
\(\frac{x^2}{16}-\frac{y^2}{b^2}=1\)
respectively satisfying \(e_1e_2=1\). If \(\alpha\) and \(\beta\) are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair \((\alpha,\beta)\) is equal to :
\(\frac{x^2}{25}+\frac{y^2}{b^2}=1\)
(\(b\lt5\)) and the hyperbola,
\(\frac{x^2}{16}-\frac{y^2}{b^2}=1\)
respectively satisfying \(e_1e_2=1\). If \(\alpha\) and \(\beta\) are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair \((\alpha,\beta)\) is equal to :
Show Hint
Write \(e_1^2=1-\frac{b^2}{25}\) and \(e_2^2=1+\frac{b^2}{16}\), use \(e_1e_2=1\) to find \(b^2\), then compute \(2ae\) for each conic.
Updated On: Nov 5, 2026
- \((8,10)\)
- \((8,12)\)
- \(\left(\frac{24}{5},10\right)\)
- \(\left(\frac{20}{3},12\right)\)
Show Solution
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The Correct Option is A
Solution and Explanation
\(e_1^2=1-\frac{b^2}{25}\), \(e_2^2=1+\frac{b^2}{16}\). \(e_1e_2=1\Rightarrow\left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1\).
Expanding: \(\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=0\Rightarrow\frac{9b^2}{400}=\frac{b^4}{400}\Rightarrow b^2=9\).
Ellipse: \(e_1=\frac45\), foci at \((\pm5\cdot\frac45,0)=(\pm4,0)\), so \(\alpha=8\).
Hyperbola: \(e_2^2=1+\frac9{16}=\frac{25}{16}\), \(e_2=\frac54\), foci at \((\pm4\cdot\frac54,0)=(\pm5,0)\), so \(\beta=10\).
Answer: (1) \((8,10)\)
Expanding: \(\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=0\Rightarrow\frac{9b^2}{400}=\frac{b^4}{400}\Rightarrow b^2=9\).
Ellipse: \(e_1=\frac45\), foci at \((\pm5\cdot\frac45,0)=(\pm4,0)\), so \(\alpha=8\).
Hyperbola: \(e_2^2=1+\frac9{16}=\frac{25}{16}\), \(e_2=\frac54\), foci at \((\pm4\cdot\frac54,0)=(\pm5,0)\), so \(\beta=10\).
Answer: (1) \((8,10)\)
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