Question

Equation of the ellipse with centre $(1,2),$ one focus at $(6,2)$ and passing through $(4,6)$ is:

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

Answer 1

The correct option is A $\frac{(x-1{)}^2}{45}+\frac{(y-2{)}^2}{20}=1$

Equation of the ellipse with centre(1,2) is given by
$\frac{(x-1{)}^2}{a^2}+\frac{(y-2{)}^2}{b^2}=1$ ... (i)

Given that focus at $(6,2)$ and centre$(1,2)$, therefore $ae=5$

We know that for ellipse $b^2=a^2(1-e^2)$ ...(ii)

Substituting the value of ae in equation (ii), we get

$b^2=a^2-a^2{e}^2$

$b^2=a^2-25$ ... (iii)

Now the ellipse passes through point(4, 6), then

$\frac{(4-1{)}^2}{a^2}+\frac{(6-2{)}^2}{b^2}=1$

$\frac{9}{a^2}+\frac{(16}{b^2}=1$

$9b^2+16a^2=a^2{b}^2$ ... (iv)

On solving equation (iii) and (iv), we get

$9(a^2-25)+16a^2=a^2(a^2-25)$

$a^4-50a^2+225=0$

$(a^2-5)(a^2-45)=0$

$a^2=5and{a}^2=45$

$a^2=5$ , will yield the negative value of b, which is not possible. then
$a^2=45and{b}^2=20$

Now substituting the value in the equation (i), we get

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

Hence, the option 'A' is correct.