Question

The equation of the ellipse with its focus at $(6,2)$, centre at $(1,2)$ and which passes through the point $(4,6)$ is?

• A. $\frac{(x-1{)}^2}{25}+\frac{(y-2{)}^2}{16}=1$
• B. $\frac{(x-1{)}^2}{25}+\frac{(y-2{)}^2}{20}=1$
• C. $\frac{(x-1{)}^2}{45}+\frac{(y-2{)}^2}{20}=1$
• D. $\frac{(x-1{)}^2}{45}+\frac{(y-2{)}^2}{16}=1$

Answer 1

Answer: (C)

$\frac{(x-1{)}^2}{45}+\frac{(y-2{)}^2}{20}=1$

Let general equation of the ellipse is $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$

where h,k are coordinates of centre of ellipse which are given as $(1,2)$

$\therefore h=1$ and $k=2$

Thus, equation of ellipse becomes,

$\frac{{(x-1)}^2}{a^2}+\frac{{(y-2)}^2}{b^2}=1$ (1)

Coordinates of focus of ellipse are given by,

$f(h+ae,k+0)$

$=f(1+ae,k)$

Given coordinates of focus are $(6,2)$

$\therefore 1+ae=6$

$\therefore ae=5$

$\therefore a^2{e}^2=25$

We know that $a^2{e}^2=c^2=a^2-b^2$

$\therefore a^2-b^2=25$

$\therefore a^2=25+b^2$

Now, ellipse passes through point $(4,6)$

$\therefore \frac{{(x-1)}^2}{25+b^2}+\frac{{(y-2)}^2}{b^2}=1$

Put $b^2=t$

$\therefore \frac{{(x-1)}^2}{25+t}+\frac{{(y-2)}^2}{t}=1$

$\therefore \frac{{(4-1)}^2}{25+t}+\frac{{(6-2)}^2}{t}=1$

$\therefore \frac{9}{25+t}+\frac{16}{t}=1$

$\therefore 9t+16(25+t)=t(25+t)$

$\therefore 9t+400+16t=25t+t^2$

$\therefore t^2-400=0$

$\therefore t^2=400$

$\therefore t=20$

$\therefore b^2=20$

$a^2=25+20$

$\therefore a^2=45$

Thus, equation of ellipse is,

$\frac{{(x-1)}^2}{45}+\frac{{(y-2)}^2}{20}=1$