Let (h,k) lie on the circle C:x^2+y^2=4 and the point (2h+1,3k+2) lie on an ellipse with e…

Question

Let $(h,k)$ lie on the circle $C:{x}^{2}+{y}^{2}=4$ and the point $(2h+1,3k+2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{{e}^{2}}$ is equal to $____$ .

Answer: 9

Accepted Answer

$(h,k)$ lie on circle ${x}^{2}+{y}^{2}=1$.

so, ${h}^{2}+{k}^{2}=1--$ (1)

$(2h+1),3k+2)$ lies on ellipse This means $x=2h+1$ and $y=3k+2$ lies on ellipse

Substitute $h=\frac{x-1}{2}$ and $k=\frac{y-2}{3}$ in equation (1)

$\begin{gathered} {(\frac{x-1}{2})}^{2}+{(\frac{y-2}{3})}^{2}=1 \ \frac{(x-1{)}^{2}}{4}+\frac{(y-2{)}^{2}}{9}=1 \end{gathered}$

This is shifted ellipse of the form $\frac{{(x-{x}_{1})}^{2}}{-2}+\frac{{(y-{y}_{1})}^{2}}{12}=1a=2,b=3$

eccentricity e $=\sqrt{1-\frac{{a}^{2}}{{b}^{2}}}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}$

${e}^{2}=\frac{5}{9}\implies \frac{5}{{e}^{2}}=9$ Ans

Solutions · 1