Thr equation of the ellipse with its centre at (1,2), one focus at (6,2) and passing through the point (4,6) is -

Dear Student,

Given that centre of ellipse is (1,2) , focus is (6,2) and passes through  (4,6)

$$\begin{gathered} \frac{(\mathrm{x}-1{)}^2}{{\mathrm{a}}^2}+\frac{(\mathrm{y}-2{)}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{This} \mathrm{is} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{ellipse}. \mathrm{We} \mathrm{need} \mathrm{to} \mathrm{find} \mathrm{a} \mathrm{and} \mathrm{b}. \\ \mathrm{Distance} \mathrm{between} \mathrm{focus} \mathrm{and} \mathrm{centre} \mathrm{is} \mathrm{c}. \\ \mathrm{So}, \mathrm{c}=\sqrt{(1-6{)}^2+(2-2{)}^2}=5 \\ \mathrm{So}, \mathrm{we} \mathrm{have} {\mathrm{a}}^2-{\mathrm{b}}^2=25 .....\left(1\right) \\ \mathrm{As}(4,6) \mathrm{lies} \mathrm{on} \mathrm{the} \mathrm{ellipse} , \mathrm{we} \mathrm{have} \\ \frac{(4-1{)}^2}{{\mathrm{a}}^2}+\frac{(6-2{)}^2}{{\mathrm{b}}^2}=1 \\ \Rightarrow 9{\mathrm{b}}^2+16{\mathrm{a}}^2={\mathrm{a}}^2{\mathrm{b}}^2 ..........\left(2\right) \\ \mathrm{Solving} \left(1\right) \mathrm{and} \left(2\right) , \mathrm{we} \mathrm{get} \\ 9{\mathrm{b}}^2+400+16{\mathrm{b}}^2=25{\mathrm{b}}^2+{\mathrm{b}}^4 \\ \Rightarrow {\mathrm{b}}^4=400 \\ \Rightarrow {\mathrm{b}}^4 - 400 = 0 \\ \Rightarrow ({\mathrm{b}}^2-20)({\mathrm{b}}^2+20)=0 \\ \Rightarrow {\mathrm{b}}^2=20 \mathrm{or} {\mathrm{b}}^2 = -20 \left(\mathrm{rejected}\right) \\ \mathrm{Now}, \mathrm{from} \left(1\right), \mathrm{we} \mathrm{get} \\ {\mathrm{a}}^2 = 20 + 25 = 45 \\ \mathrm{The} \mathrm{equation} \mathrm{of} \mathrm{ellipse} \mathrm{is} \\ \frac{(\mathrm{x}-1{)}^2}{45}+\frac{(\mathrm{y}-2{)}^2}{20}=1 \end{gathered}$$

So option a is correct.

Regards