Question

The sum of the squares of the eccentricities of the conics $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$ is (a) 2 (b) $\sqrt{2}$ (c) $\frac{\sqrt{7}}{\sqrt{3}}$ (d) $\sqrt{7}$

          The sum of the squares of the eccentricities of the conics $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$ is
(a) 2
(b) $\sqrt{2}$
(c) $\frac{\sqrt{7}}{\sqrt{3}}$
(d) $\sqrt{7}$

Added by Gregorio S.

Geometry A Common Core Curriculum

Ron Larson, Laurie Boswell 1st Edition

Instant Answer

Solved by Expert Darshan Maheshwari

02/01/2022

Step 1

Step 1: Find the eccentricity of the ellipse: Given equation of ellipse: $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ $a^2 = 4$ and $b^2 = 3$ Using the formula $b^2 = a^2(1 - e^2)$, we can find $e^2$: $3 = 4(1 - e^2)$ $3 = 4 - 4e^2$ $4e^2 = 1$ $e^2 = \frac{1}{4}$ $e = Show more…

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The sum of the squares of the eccentricities of the conics $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$ is (a) 2 (b) $\sqrt{2}$ (c) $\frac{\sqrt{7}}{\sqrt{3}}$ (d) $\sqrt{7}$