Let the eccentricity of Ellipse, x2a2+y2b2=1,(a>b) is 14. If the Ellipse passes through (−425,3). Then a2+b2=________

For e=14,b2=15a216 and ‘P’ lies on Ellipse ⇒325a2+9b2=1.Solving b2=15,a2=16. Sum = 31

Let the eccentricity of Ellipse, x2a2+y2b2=1,(a>b) is 14. If the Ellipse passes through (−425,3). Then a2+b2=________

For e=14,b2=15a216 and ‘P’ lies on Ellipse ⇒325a2+9b2=1.Solving b2=15,a2=16. Sum = 31

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Let the eccentricity of Ellipse, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ is $\frac{1}{4}$ . If the Ellipse passes through $(- 4 \sqrt{\frac{2}{5}}, 3)$ . Then $a^{2} + b^{2} =________$

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For $e = \frac{1}{4}, b^{2} = \frac{15 a^{2}}{16}$ and ‘P’ lies on Ellipse $\Rightarrow \frac{32}{5 a^{2}} + \frac{9}{b^{2}} = 1$ .
Solving $b^{2} = 15, a^{2} = 16$ . Sum = 31

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Let the eccentricity of Ellipse, x2a2+y2b2=1,(a>b) is 14. If the Ellipse passes through (−425,3). Then a2+b2=________

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Let the eccentricity of Ellipse, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ is $\frac{1}{4}$ . If the Ellipse passes through $(- 4 \sqrt{\frac{2}{5}}, 3)$ . Then $a^{2} + b^{2} =________$