If e and e' are the eccentricities of the ell | vedclass

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

Question

(D) For the ellipse $5x^2 + 9y^2 = 45$,we divide by $45$ to get $\frac{x^2}{9} + \frac{y^2}{5} = 1$. Here $a^2 = 9$ and $b^2 = 5$. The eccentricity $e

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) For the ellipse $5x^2 + 9y^2 = 45$,we divide by $45$ to get $\frac{x^2}{9} + \frac{y^2}{5} = 1$. Here $a^2 = 9$ and $b^2 = 5$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$. For the hyperbola $5x^2 - 4y^2 = 45$,we divide by $45$ to get $\frac{x^2}{9} - \frac{y^2}{45/4} = 1$. Here $a^2 = 9$ and $b^2 = \frac{45}{4}$. The eccentricity $e'$ is given by $e' = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{45/4}{9}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$. Therefore,$ee' = \frac{2}{3} 	imes \frac{3}{2} = 1$.

If $e$ and $e'$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee' = $

Similar Questions

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be an ellipse,whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

The equation of the line passing through the points of intersection of the parabola $x^2 = 8y$ and the ellipse $\frac{x^2}{3} + y^2 = 1$ is

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then,the value of $b^2$ is

An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and the parabola $x^2 = 4(y + b)$ are such that the two foci of the ellipse and the end points of the latus rectum of the parabola are the vertices of a square. The eccentricity of the ellipse is

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. The equation of the common tangent with a positive slope to the circle and the hyperbola is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module