Equations of the latus rectum of the ellipse | vedclass

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

Question

(C) The given equation of the ellipse is $9x^2 + 4y^2 - 18x - 8y - 23 = 0$. Rearranging the terms,we get $9(x^2 - 2x) + 4(y^2 - 2y) = 23$. Completing

Vedclass · Accepted Answer

$y = 1 \pm \sqrt{5}$

Vedclass · Answer

(C) The given equation of the ellipse is $9x^2 + 4y^2 - 18x - 8y - 23 = 0$. Rearranging the terms,we get $9(x^2 - 2x) + 4(y^2 - 2y) = 23$. Completing the square,$9(x^2 - 2x + 1) + 4(y^2 - 2y + 1) = 23 + 9 + 4$. $9(x - 1)^2 + 4(y - 1)^2 = 36$. Dividing by $36$,we get $\frac{(x - 1)^2}{4} + \frac{(y - 1)^2}{9} = 1$. Here,$a^2 = 4$ and $b^2 = 9$,so $a The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$. The equations of the latus rectum for a vertical ellipse are $y - k = \pm be$,where $(h, k) = (1, 1)$. $y - 1 = \pm 3 	imes \frac{\sqrt{5}}{3} = \pm \sqrt{5}$. Therefore,$y = 1 \pm \sqrt{5}$.

Equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:

Similar Questions

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

Find the equation of the ellipse whose vertices are $(\pm 13, 0)$ and foci are $(\pm 5, 0)$.

The length of the latus rectum of $9x^2 + 25y^2 - 90x - 150y + 225 = 0$ is

The ellipse $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ circumscribes the rectangle $R$ and passes through the point $(0, 4)$. What is the eccentricity of the ellipse $E_2$?

The conic represented by $x = 2(\cos t + \sin t), y = 5(\cos t - \sin t)$ is .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module