An ellipse E: frac{x^{2}}{a^{2}}+frac{y^{2}}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The hyperbola $H$ is given by $\frac{y^{2}}{64} - \frac{x^{2}}{49} = 1$. Its vertices are $(0, \pm 8)$.Since the ellipse $E: \frac{x^{2}}{a^{2}} +

Vedclass · Accepted Answer

$1552$

Vedclass · Answer

(B) The hyperbola $H$ is given by $\frac{y^{2}}{64} - \frac{x^{2}}{49} = 1$. Its vertices are $(0, \pm 8)$.Since the ellipse $E: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ passes through $(0, \pm 8)$,we have $b^2 = 64$,so $b = 8$.The eccentricity of the hyperbola $H$ is $e_H = \sqrt{1 + \frac{49}{64}} = \sqrt{\frac{113}{64}} = \frac{\sqrt{113}}{8}$.The eccentricity of the ellipse $E$ is $e_E = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{a^2}{64}}$.Given $e_E 	imes e_H = \frac{1}{2}$,we have $\sqrt{1 - \frac{a^2}{64}} 	imes \frac{\sqrt{113}}{8} = \frac{1}{2}$.Squaring both sides: $(1 - \frac{a^2}{64}) 	imes \frac{113}{64} = \frac{1}{4}$ $\Rightarrow 1 - \frac{a^2}{64} = \frac{64}{4 	imes 113} = \frac{16}{113}$.Thus,$\frac{a^2}{64} = 1 - \frac{16}{113} = \frac{97}{113}$,so $a^2 = \frac{64 	imes 97}{113}$.The length of the latus rectum $l = \frac{2a^2}{b} = \frac{2}{8} 	imes \frac{64 	imes 97}{113} = \frac{16 	imes 97}{113} = \frac{1552}{113}$.Therefore,$113l = 1552$.

Similar Questions

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

The common tangents to the circle $x^2+y^2=2$ and the parabola $y^2=8x$ touch the circle at the points $P, Q$ and the parabola at the points $R, S$. Then the area of the quadrilateral $PQRS$ is

The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to:

The angle between the curves $2x^2 + y^2 = 20$ and $4y^2 - x^2 = 8$ at a point where they intersect in the $4^{th}$ quadrant is

The number of points of intersection of $|z - (4 + 3i)| = 2$ and $|z| + |z - 4| = 6$,$z \in \mathbb{C}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module