Let the length of a latus rectum of an ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\) be \(10\). If its eccentricity is the minimum value of the function \(f(\mathrm{t})={\mathrm{t}}^{2}+\mathrm{t}+\frac{11}{12}\), \(\mathrm{t}\in R\), then \({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\) is equal to :

[JEE Main 2025, 7 Apr (Shift 2)]

Question

Let the length of a latus rectum of an ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\) be \(10\). If its eccentricity is the minimum value of the function \(f(\mathrm{t})={\mathrm{t}}^{2}+\mathrm{t}+\frac{11}{12}\), \(\mathrm{t}\in R\), then \({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\) is equal to :

[JEE Main 2025, 7 Apr (Shift 2)]

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Let the length of a latus rectum of an ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\) be \(10\). If its e…

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