Question

Equation of the ellipse with vertices $(0,\pm 17)$ and foci $(0,\pm 8)$ is:

• A. $\frac{x^2}{289}+\frac{y^2}{225}=1$
• B. $\frac{x^2}{225}+\frac{y^2}{289}=1$
• C. $\frac{x^2}{132}+\frac{y^2}{289}=1$
• D. $\frac{x^2}{196}+\frac{y^2}{289}=1$

Answer 1

Answer: (B)

$\frac{x^2}{225}+\frac{y^2}{289}=1$

Given,Vertices $(0,\underset{–}{+}17)$ and foci $(0,\underset{–}{+}8)$,so center of ellipse will be $(0,0)$
As foci are on y-axis,major axis will be y-axis
Let ,$2b$ and $2a$ be lengths of major and minor axis respectively and $e$ be eccentricity of ellipse.
Now ,length of major axis will be form vertices,$2b=34$
We know that ,distance between foci is $2be$,then $e=\frac{8}{17}$
and also we know that $b^2(1-e^2)=a^2⟹a^2=225$
Then equation of ellipse will be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1⟹\frac{x^2}{225}+\frac{y^2}{289}=1$