Question

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

Answer 1

Answer: (C)

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

We know that,

The equation of ellipse whose centre $(h,k)$.

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

Given centre of ellipse $(h,k)=(1,2)$

Then, equation of ellipse $\frac{{(x-1)}^2}{a^2}+\frac{{(y-2)}^2}{b^2}=1$ ……. (1)

But, the ellipse passes through given point $(x,y)=(4,6)$

By equation (1), we get

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

$\frac{3^2}{a^2}+\frac{4^2}{b^2}=1$

$\Rightarrow 9b^2+16a^2=a^2{b}^2$

$\Rightarrow 16a^2+9b^2=a^2{b}^2$ …… (2)

Now, distance between focus and centre is $c=\sqrt{a^2-b^2}$

So,

$c=\sqrt{{(1-6)}^2+{(2-2)}^2}$

$c=\sqrt{5^2}$

$c=5$

$\sqrt{a^2-b^2}=5$

On squaring both sides, we get,

$a^2-b^2=25$ …… (3)

By equation (2) and (3), we get

$9b^2+400+16b^2=25b^2+b^4$

$25b^2+400=25b^2+b^4$

$400=b^4$

$b^4-400=0$

$(b^2-20)(b^2+20)=0$

$b^2-20=0$ and $b^2+20=0$

$b^2=20$ and $b^2=-20$ (Rejected)

Put the value of $b^2$ in equation (3) and we get,

$a^2-b^2=25$

$a^2-20=25$

$a^2=45$

Now, put the value of $a^2$ and $b^2$ in equation (1), we get,

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