Let the product of the focal distances of the point \(\left(\sqrt{3},\frac{1}{2}\right)\text{ on the ellipse }\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,(\mathrm{a}>\mathrm{b})\text{, be }\frac{7}{4}.\) Then the absolute difference of the eccentricities of two such ellipses is

[JEE Main 2025, 24 Jan (Shift 1)]

Question

Let the product of the focal distances of the point \(\left(\sqrt{3},\frac{1}{2}\right)\text{ on the ellipse }\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,(\mathrm{a}>\mathrm{b})\text{, be }\frac{7}{4}.\) Then the absolute difference of the eccentricities of two such ellipses is

[JEE Main 2025, 24 Jan (Shift 1)]

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Let the product of the focal distances of the point \(\left(\sqrt{3},\frac{1}{2}\right)\text{ on the ellipse }\frac{{\ma…

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