Q.11. The eccentricity of an ellipse E with centre at the origin O is and its directrices are . Let

Q: What is the correct answer for JEE Main Mathematics Q11?

The correct answer is D. Sol. For ellipse E: Eccentricity $e = \frac{\sqrt{3}}{2}$. Directrices are $x = \pm \frac{a}{e} = \pm \frac{4\sqrt{6}}{3}$. So $\frac{a}{\sqrt{3}/2} = \frac{4\sqrt{6}}{3}$. $a = \frac{\sqrt{3}}{2} \cdot \frac{4\sqrt{6}}{3} = \frac{2\sqrt{18}}{3} = \frac{2 \cdot 3\sqrt{2}}{3} = 2\sqrt{2}$. For ellipse, $b^2 = a^2(1-e^2) = (2\sqrt{2})^2(1 - (\frac{\sqrt{3}}{2})^2) = 8(1 - \frac{3}{4}) = 8(\frac{1}{4

Q: Q.11. The eccentricity of an ellipse E with centre at the origin O is and its directrices are . Let H: be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose…

The correct answer is D. Sol. For ellipse E: Eccentricity $e = \frac{\sqrt{3}}{2}$. Directrices are $x = \pm \frac{a}{e} = \pm \frac{4\sqrt{6}}{3}$. So $\frac{a}{\sqrt{3}/2} = \frac{4\sqrt{6}}{3}$. $a = \frac{\sqrt{3}}{2} \cdot \frac{4\sqrt{6}}{3} = \frac{2\sqrt{18}}{3} = \frac{2 \cdot 3\sqrt{2}}{3} = 2\sqrt{2}$. For ellipse, $b^2 = a^2(1-e^2) = (2\sqrt{2})^2(1 - (\frac{\sqrt{3}}{2})^2) = 8(1 - \frac{3}{4}) = 8(\frac{1}{4

Q: What are the options for this JEE Main Mathematics question?

A. $\frac{4\sqrt{2}}{7}$ · B. $\frac{4}{7}$ · C. $\frac{8\sqrt{2}}{7}$ · D. $\frac{8}{7}$

Question

Q.11. The eccentricity of an ellipse E with centre at the origin O is  and its directrices are . Let H:  be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is :

FirstInTest · Accepted Answer

Sol. For ellipse E: Eccentricity $e = \frac{\sqrt{3}}{2}$. Directrices are $x = \pm \frac{a}{e} = \pm \frac{4\sqrt{6}}{3}$. So $\frac{a}{\sqrt{3}/2} = \frac{4\sqrt{6}}{3}$. $a = \frac{\sqrt{3}}{2} \cdot \frac{4\sqrt{6}}{3} = \frac{2\sqrt{18}}{3} = \frac{2 \cdot 3\sqrt{2}}{3} = 2\sqrt{2}$. For ellipse, $b^2 = a^2(1-e^2) = (2\sqrt{2})^2(1 - (\frac{\sqrt{3}}{2})^2) = 8(1 - \frac{3}{4}) = 8(\frac{1}{4}) = 2$. So $b = \sqrt{2}$. Semi-major axis of E is $a_E = 2\sqrt{2}$. Minor axis of E is $2b_E = 2\sqrt{2}$. For hyperbola H: Eccentricity $e_H = a_E = 2\sqrt{2}$. Length of latus rectum of H is $\frac{2b_H^2}{a_H}$. This is equal to the length of minor axis of E, which is $2b_E = 2\sqrt{2}$. So $\frac{2b_H^2}{a_H} = 2\sqrt{2}$. $b_H^2 = \sqrt{2}a_H$. For hyperbola, $b_H^2 = a_H^2(e_H^2-1)$. So $\sqrt{2}a_H = a_H^2((2\sqrt{2})^2-1) = a_H^2(8-1) = 7a_H^2$. Since $a_H 
e 0$, we can divide by $a_H$: $\sqrt{2} = 7a_H$. So $a_H = \frac{\sqrt{2}}{7}$. Then $b_H^2 = \sqrt{2}a_H = \sqrt{2} \cdot \frac{\sqrt{2}}{7} = \frac{2}{7}$. The distance between the foci of H is $2a_H e_H = 2 \cdot \frac{\sqrt{2}}{7} \cdot 2\sqrt{2} = 2 \cdot \frac{4}{7} = \frac{8}{7}$.

Answer

$\frac{4\sqrt{2}}{7}$

Answer

$\frac{4}{7}$

Answer

$\frac{8\sqrt{2}}{7}$

Answer

$\frac{8}{7}$

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