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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

Let the eccentricity of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b\), be \(\frac{1}{4}\). If this ellipse passes through the point \(\left(-4 \sqrt{\frac{2}{5}}, 3\right)\), then \(a^{2}+b^{2}\) is equal to

  1. A \(31\)
  2. B \(29\)
  3. C \(32\)
  4. D \(34\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(31\)

Step-by-step Solution

Detailed explanation

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 a > b\) \(e^{2}=1-\frac{b^{2}}{a^{2}}\) \(\frac{1}{16}=1-\frac{b^{2}}{a^{2}}\) \(\frac{b^{2}}{a^{2}}=1-\frac{1}{16}=\frac{15}{16} \Rightarrow b^{2}=\frac{15}{16} a^{2}\) \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)…