Question

Find the equation of the ellipse with its centre at $(4,-1)$ focus at $(1,-1)$ and given that it passes through $(8,0).$

Answer 1

Let C be the centre and F and F' be the foci of the ellipse.
Given $F(1,-1),C=(4,-1)$
$F^{\prime }(x,y)$ (say)
C is the Mid-point of FF',
$4=\frac{x+1}{2},-1=\frac{y-1}{2}$
$\Rightarrow x=7,y=-1$
$F^{\prime }(7,-1)$

Since, the y-coordinates of F and F' are equal therefore, the transverse axis of the ellipse is parallel to x-axis.

The equation of the ellipse with centre at (4, -1) is
$\frac{(x-4{)}^2}{a^2}+\frac{(y+1{)}^2}{b^2}=1$ ......... (i)
$|CF|=\sqrt{(4-1{)}^2+(-1+1{)}^2}$
$|CF|=3$
$\Rightarrow ae=3$
$\Rightarrow a^2{e}^2=9$

Now, $b^2=a^2-a^2{e}^2$
$\Rightarrow b^2=a^2-9$ ......... (ii)

As ellipse (i) passes through (8, 0),
$\frac{16}{a^2}+\frac{1}{b^2}=1$
$\Rightarrow 16b^2+a^2=a^2{b}^2$ ........ (iii)
$16(a^2-9)+a^2(a^2-9)$
$\Rightarrow a^4-16a^2+144=0$
$\Rightarrow a^4-18a^2-8a^2+144=0$
$\Rightarrow (a^2-18)(a^2-8)=0$
$\Rightarrow a^2=18,a^2=8$

$a^2=8$ is inadmissible as it yields $b^2$ a negative value.
$\therefore a^2=18$
$\therefore$ From (ii), $b^2=9$
$\therefore$ Equation of the ellipse is
$\frac{(x-4{)}^2}{18}+\frac{(y+1{)}^2}{9}=1$
$\Rightarrow (x-4{)}^2+(y+1{)}^2=18$
$x^2-8x+16+2y^2+4y+2=18$
$\Rightarrow x^2+2y^2-8x+4y=0$