Planet M orbits around its sun,S,in an ellipt | vedclass

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

Question

(B) Let the major axis be along the $y$-axis. The distance between the closest point and the farthest point is the length of the major axis,$2b = 2 +

Vedclass · Accepted Answer

$\frac{x^2}{36} + \frac{(y + 8)^2}{100} = 1$

Vedclass · Answer

(B) Let the major axis be along the $y$-axis. The distance between the closest point and the farthest point is the length of the major axis,$2b = 2 + 18 = 20$,so $b = 10$.The center of the ellipse is at $(0, -8)$ because the distance from the focus $(0, 0)$ to the closest point $(0, 2)$ is $2$,and to the farthest point $(0, -18)$ is $18$. The center is the midpoint of the major axis,which is $(0, \frac{2 + (-18)}{2}) = (0, -8)$.The distance from the center $(0, -8)$ to the focus $(0, 0)$ is $c = 8$.Using $c^2 = b^2 - a^2$,we have $8^2 = 10^2 - a^2$,so $64 = 100 - a^2$,which gives $a^2 = 36$,so $a = 6$.The equation of the ellipse with center $(h, k) = (0, -8)$ and major axis along the $y$-axis is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.Substituting the values,we get $\frac{x^2}{36} + \frac{(y + 8)^2}{100} = 1$.

Similar Questions

The eccentricity of the ellipse $(x-3)^2 + (y-4)^2 = \frac{y^2}{9} + 16$ is -

When does the line $x \cos \alpha + y \sin \alpha = p$ touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$?

Let $P$ be any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If $S_1$ and $S_2$ are its foci,then the maximum area of $\Delta PS_1S_2$ is (in square units):

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut each other orthogonally, then $a^2-b^2$ equals to

Vedclass Test Series

Exam Paper Generator

Online Exam Module