The eccentricity of the ellipse left( frac{x | vedclass

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

Question

(D) Given equation: $\left( \frac{x - 3}{y} \right)^2 + \left( \frac{y - 4}{y} \right)^2 = \frac{1}{9}$$(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$$9(x - 3

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) Given equation: $\left( \frac{x - 3}{y} \right)^2 + \left( \frac{y - 4}{y} \right)^2 = \frac{1}{9}$$(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$$9(x - 3)^2 + 9(y^2 - 8y + 16) = y^2$$9(x - 3)^2 + 8y^2 - 72y + 144 = 0$$9(x - 3)^2 + 8(y^2 - 9y) = -144$$9(x - 3)^2 + 8(y - 4.5)^2 = -144 + 8(20.25) = 18$$\frac{(x - 3)^2}{2} + \frac{(y - 4.5)^2}{18/8} = 1$$\frac{(x - 3)^2}{2} + \frac{(y - 4.5)^2}{2.25} = 1$Here,$a^2 = 2.25$ and $b^2 = 2$.Since $a^2 > b^2$,the eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2}{2.25} = 1 - \frac{2}{9/4} = 1 - \frac{8}{9} = \frac{1}{9}$.Therefore,$e = \frac{1}{3}$.

The eccentricity of the ellipse $\left( \frac{x - 3}{y} \right)^2 + \left( 1 - \frac{4}{y} \right)^2 = \frac{1}{9}$ is

Similar Questions

The equation of the ellipse passing through the origin and having foci at $(1, 0)$ and $(3, 0)$ is .....

The eccentricity of the ellipse $(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$ is

If the line $y = mx + c$ touches the ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,then $c = $

The length of the minor axis (along $y$-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}$. If this ellipse touches the line $x+6y=8$,then its eccentricity is

If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse,then the eccentricity of the ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module