The equations of the directrices of the ellip | vedclass

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

Question

(A) Given equation: $9x^2 + 4y^2 - 18x - 16y - 11 = 0$ Rearranging terms: $9(x^2 - 2x) + 4(y^2 - 4y) = 11$ Completing the square: $9(x-1)^2 - 9 + 4(y-

Vedclass · Accepted Answer

$y = 2 \pm \frac{9}{\sqrt{5}}$

Vedclass · Answer

(A) Given equation: $9x^2 + 4y^2 - 18x - 16y - 11 = 0$ Rearranging terms: $9(x^2 - 2x) + 4(y^2 - 4y) = 11$ Completing the square: $9(x-1)^2 - 9 + 4(y-2)^2 - 16 = 11$ $9(x-1)^2 + 4(y-2)^2 = 36$ Dividing by $36$: $\frac{(x-1)^2}{4} + \frac{(y-2)^2}{9} = 1$ Here,$a^2 = 4$ and $b^2 = 9$. Since $b^2 > a^2$,the major axis is vertical. Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$ The center is $(h, k) = (1, 2)$. The equations of the directrices for a vertical ellipse are $y - k = \pm \frac{b}{e}$. $y - 2 = \pm \frac{3}{\sqrt{5}/3} = \pm \frac{9}{\sqrt{5}}$ $y = 2 \pm \frac{9}{\sqrt{5}}$

The equations of the directrices of the ellipse $9x^2 + 4y^2 - 18x - 16y - 11 = 0$ are

Similar Questions

The equation of an ellipse,whose vertices are $(2, -2)$ and $(2, 4)$ and eccentricity is $\frac{1}{3}$,is

If the tangent to the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is parallel to the $x$-axis,then what is the point of tangency?

If the line $x \cos \alpha + y \sin \alpha = p$ is normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then

If a tangent of slope $2$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ touches the circle $x^2+y^2=4$,then the maximum value of $ab$ is

Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci,then the square of the sum of the lengths of its minor and major axes is equal to $...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module