If the points of intersection of the ellipses x^2+2 y^2-6 x-12 y+23=0 and 4 x^2+2 y^2-20 x…

Question

If the points of intersection of the ellipses $x^2+2 y^2-6 x-12 y+23=0$ and 
$4 x^2+2 y^2-20 x-12 y+35=0$ lie on a circle of radius $r$ and centre $(a, b)$, then the 
value of $a b+18 r^2$ is :

Accepted Answer

Let the given ellipses be :

$\begin{gathered} {S}_{1}\equiv {x}^{2}+2{y}^{2}-6x-12y+23=0 \ {S}_{2}\equiv 4{x}^{2}+2{y}^{2}-20x-12y+35=0 \end{gathered}$

The equation of a curve passing through the intersection of these two ellipses is given by the family of curves ${S}_{1}+\lambda {S}_{2}=0$ :

$({x}^{2}+2{y}^{2}-6x-12y+23)+\lambda (4{x}^{2}+2{y}^{2}-20x-12y+35)=0$

$\Rightarrow$$(1+4\lambda ){x}^{2}+(2+2\lambda ){y}^{2}-(6+20\lambda )x-(12+12\lambda )y+(23+35\lambda )=0$

Condition for circle: coefficient of ${x}^{2}=$ coefficient of ${y}^{2}$

$1+4\lambda =2+2\lambda $

$\Rightarrow$ $2\lambda =1$

$\Rightarrow$ $\lambda =\frac{1}{2}$

Substitute $\lambda =\frac{1}{2}$ :

$3{x}^{2}+3{y}^{2}-16x-18y+\frac{81}{2}=0$

$\Rightarrow$ ${x}^{2}+{y}^{2}-\frac{16}{3}x-6y+\frac{27}{2}=0$

$\begin{gathered} 2g=-\frac{16}{3}\implies g=-\frac{8}{3} \ 2f=-6\implies f=-3 \ (a,b)=(-g,-f)=(\frac{8}{3},3) \ a=\frac{8}{3},b=3 \end{gathered}$

$\begin{gathered} {r}^{2}={g}^{2}+{f}^{2}-c \ {r}^{2}={(-\frac{8}{3})}^{2}+(-3{)}^{2}-\frac{27}{2} \ {r}^{2}=\frac{64}{9}+9-\frac{27}{2} \ {r}^{2}=\frac{128+162-243}{18} \ {r}^{2}=\frac{47}{18} \end{gathered}$

$\begin{gathered} ab+18{r}^{2}=(\frac{8}{3})(3)+18(\frac{47}{18}) \ ab+18{r}^{2}=8+47=55 \end{gathered}$

Solutions · 1