Question:
If a hyperbola has length of its transverse axis \(=\sqrt{2}\) and it has the same foci as that of the ellipse \(4x^{2}+3y^{2}=24\), then this hyperbola also passes through the point whose coordinates are :
If a hyperbola has length of its transverse axis \(=\sqrt{2}\) and it has the same foci as that of the ellipse \(4x^{2}+3y^{2}=24\), then this hyperbola also passes through the point whose coordinates are :
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Find the foci of the ellipse, get the hyperbola from \(2a=\sqrt{2}\) and \(c^{2}=a^{2}+b^{2}\), then test the points.
Updated On: Nov 12, 2026
- \(\left(2\sqrt{3},\frac{3}{\sqrt{2}}\right)\)
- \(\left(2\sqrt{3},\frac{3}{2}\right)\)
- \(\left(\frac{1}{\sqrt{6}},1\right)\)
- \(\left(\frac{1}{\sqrt{6}},\frac{\sqrt{2}}{3}\right)\)
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The Correct Option is A
Solution and Explanation
The ellipse is \(\frac{x^{2}}{6}+\frac{y^{2}}{8}=1\). Here \(a^{2}=8\) (along y), \(b^{2}=6\), so \(c^{2}=8-6=2\) and the foci are \((0,\pm\sqrt{2})\).
The hyperbola has the same foci, so its transverse axis is along the y-axis: \(\frac{y^{2}}{A^{2}}-\frac{x^{2}}{B^{2}}=1\) with \(C^{2}=2\).
Transverse axis \(2A=\sqrt{2}\Rightarrow A^{2}=\frac{1}{2}\); \(B^{2}=C^{2}-A^{2}=2-\frac{1}{2}=\frac{3}{2}\).
Hyperbola: \(2y^{2}-\frac{2x^{2}}{3}=1\).
Test \(\left(2\sqrt{3},\frac{3}{\sqrt{2}}\right)\): \(2\cdot\frac{9}{2}-\frac{2\cdot12}{3}=9-8=1\). Satisfied.
Test \(\left(2\sqrt{3},\frac{3}{2}\right)\): \(2\cdot\frac{9}{4}-8=-\frac{7}{2}\ne1\). Test \(\left(\frac{1}{\sqrt{6}},1\right)\): \(2-\frac{2}{18}\ne1\). Test \(\left(\frac{1}{\sqrt{6}},\frac{\sqrt{2}}{3}\right)\): \(\frac{4}{9}-\frac{1}{9}=\frac{1}{3}\ne1\).
Answer: (1) \(\left(2\sqrt{3},\frac{3}{\sqrt{2}}\right)\)
The hyperbola has the same foci, so its transverse axis is along the y-axis: \(\frac{y^{2}}{A^{2}}-\frac{x^{2}}{B^{2}}=1\) with \(C^{2}=2\).
Transverse axis \(2A=\sqrt{2}\Rightarrow A^{2}=\frac{1}{2}\); \(B^{2}=C^{2}-A^{2}=2-\frac{1}{2}=\frac{3}{2}\).
Hyperbola: \(2y^{2}-\frac{2x^{2}}{3}=1\).
Test \(\left(2\sqrt{3},\frac{3}{\sqrt{2}}\right)\): \(2\cdot\frac{9}{2}-\frac{2\cdot12}{3}=9-8=1\). Satisfied.
Test \(\left(2\sqrt{3},\frac{3}{2}\right)\): \(2\cdot\frac{9}{4}-8=-\frac{7}{2}\ne1\). Test \(\left(\frac{1}{\sqrt{6}},1\right)\): \(2-\frac{2}{18}\ne1\). Test \(\left(\frac{1}{\sqrt{6}},\frac{\sqrt{2}}{3}\right)\): \(\frac{4}{9}-\frac{1}{9}=\frac{1}{3}\ne1\).
Answer: (1) \(\left(2\sqrt{3},\frac{3}{\sqrt{2}}\right)\)
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