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

How many real tangents can be drawn to the ellipse $5x^2 + 9y^2 = 32$ from the point $(2, 3)$?

If the latus rectum of an ellipse is half of its minor axis,then its eccentricity is ...

Let $L$ be the distance between two parallel normals of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Then the maximum value of $L$ is:

If the equations $x = 1 + 2 \cos \theta$ and $y = 2 + \sin \theta$ for $0 \leq \theta < 2 \pi$ represent an ellipse,then the point of intersection of the normal drawn at $P(\theta = \pi/4)$ to this ellipse and its major axis is:

The eccentricity of an ellipse passing through $(3 \sqrt{2}, \sqrt{10})$ with foci at $(-4,0)$ and $(4,0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module