If the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ meets the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ on the $x$-axis and the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ on the $y$-axis, then the eccentricity of the ellipse is :
Solution
<p>${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ meets the line ${x \over 7} + {y \over {2\sqrt 6 }} = 1$ on the x-axis</p>
<p>So, $a = 7$</p>
<p>and ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ meets the line ${x \over 7} - {y \over {2\sqrt 6 }} = 1$ on the y-axis</p>
<p>So, $b = 2\sqrt 6$</p>
<p>Therefore, ${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {{24} \over {49}}$</p>
<p>$e = {5 \over 7}$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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