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

Answer 1

Answer: (D)

Given, Focus $S=(6,2)$
Centre $C=(1,2)=(h,k)$
Point $P=(4,6)$
Required Equation of ellipse is
$\frac{(x-1{)}^2}{a^2}+\frac{(y-2{)}^2}{b^2}=1\dots$ (i)
Since, Eq. (i) passes through $P(4,6)$
$\frac{(4-1{)}^2}{a^2}+\frac{(6-2{)}^2}{b^2}=1$
$\frac{9}{a^2}+\frac{16}{b^2}=1\dots .$ (ii)
Since, Focus $=(6,2)$
$(h+ae,k)=(6,2)$
$\therefore h+ae=6$
$k=2$
$1+ae=6$
$ae=5$
$a^2{e}^2=25$
$b^2=a^2\left(1-e^2\right)$
$b^2=a^2-a^2{e}^2$
$b^2=a^2-25$
$a^2=b^2+25\dots$ (iii)
Put $a^2$ value in Eq. (ii),
$\frac{9}{b^2+25}+\frac{16}{b^2}=1$
$9b^2+16\left(b^2+25\right)=b^2\left(b^2+25\right)$
$9b^2+16b^2+400=b^4+25b^2$
$b^4=400$
$b^2=20$
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 $1$]