JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
For the hyperbola \(H : x ^{2}- y ^{2}=1\) and the ellipse \(E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b>0\), let the \((1)\) eccentricity of \(E\) be reciprocal of the eccentricity of \(H\), and \((2)\) the line \(y=\sqrt{\frac{5}{2}} x+K\) be a common tangent of \(E\) and \(H\) Then \(4\left(a^{2}+b^{2}\right)\) is equal to
- A \(2\)
- B \(0\)
- C \(1\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(e _{ E }=\sqrt{1-\frac{ b ^{2}}{ a ^{2}}}, e _{ H }=\sqrt{2}\) If \(\Rightarrow e _{ E }=\frac{1}{ e _{ H }}\) \(\Rightarrow \frac{ a ^{2}- b ^{2}}{ a ^{2}}=\frac{1}{2}\) \(2 a ^{2-2 b } 2= a ^{2}\) \(a ^{2}=2 b ^{2}\) and \(y =\sqrt{\frac{5}{2}} x + k\) is tangent to ellipse…
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