Given, Focus $S=(6,2)$
Centre $C=(1,2)=(h, k)$
Point $P=(4,6)$
Required Equation of ellipse is
$\frac{(x-1)^{2}}{a^{2}}+\frac{(y-2)^{2}}{b^{2}}=1 \ldots$ (i)
Since, Eq. (i) passes through $P(4,6)$
$\frac{(4-1)^{2}}{a^{2}}+\frac{(6-2)^{2}}{b^{2}}=1$
$\frac{9}{a^{2}}+\frac{16}{b^{2}}=1 \ldots .$ (ii)
Since, Focus $=(6,2)$
$(h +a e, k)=(6,2)$
$\therefore h +a e=6$
$k=2$
$1+a e=6$
$a e=5$
$a^{2} e^{2}=25$
$b^{2}=a^{2}\left(1-e^{2}\right)$
$b^{2}=a^{2}-a^{2} e^{2}$
$b^{2}=a^{2}-25$
$a^{2}=b^{2}+25 \ldots$ (iii)
Put $a^{2}$ value in Eq. (ii),
$\frac{9}{b^{2}+25}+\frac{16}{b^{2}}=1$
$9 b^{2}+16\left(b^{2}+25\right)=b^{2}\left(b^{2}+25\right)$
$9 b^{2}+16 b^{2}+400=b^{4}+25 b^{2}$
$b^{4}=400$
$b^{2}=20$
From Eq. (iii),
$a^{2}=20+25$
$a^{2}=45$
Put $a^{2}, b^{2}$ Value in Eq. (i),
$\frac{(x-1)^{2}}{45}+\frac{(y-2)^{2}}{20}=1$
$[\therefore $ Answer written in the paper was wrong in RHS it should be $1$]