JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 9)
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 :
$$\frac{5}{7}$$
$$\frac{2 \sqrt{6}}{7}$$
$$\frac{3}{7}$$
$$\frac{2 \sqrt{5}}{7}$$
Explanation
$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ meets the line $${x \over 7} + {y \over {2\sqrt 6 }} = 1$$ on the x-axis
So, $$a = 7$$
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
So, $$b = 2\sqrt 6 $$
Therefore, $${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {{24} \over {49}}$$
$$e = {5 \over 7}$$
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