# Thread: Find equation of ellpise

1. ## Find equation of ellpise

Use the locus definition of an ellipse to find the equation of the ellipse with focus (3,4), corresponding directrix line x+y=1 and eccentricity 1/2.

so e=1/2, one focus(3,4)
let the center be (c,d)
then half the length of the major axis=a
distance of focus from center=ae distance of directrix from center =a/e.
center is on the line y=x+1
half distance between focus and directrix =$\displaystyle \frac{1}{2}\frac{3+4-1}{\sqrt{2}}=\frac{3\sqrt2}{2}$

from ae, $\displaystyle \frac{1}{2}a=\sqrt{(c-3)^2+(d-4)^2}$
$\displaystyle a=2\sqrt{(c-3)^2+(d-4)^2}$
from a/e, $\displaystyle 2a=\frac{c+d-1}{\sqrt{2}}$ finding the length of a perpendicular from a point to a line.
so,$\displaystyle 4\sqrt{(c-3)^2+(d-4)^2}=\frac{c+d-1}{\sqrt{2}}$
i work it out to get $\displaystyle 31c^2+31d^2-98c-258d-2cd+799=0$
also $\displaystyle a=\frac{c+d-1}{\sqrt{2}}-\frac{3\sqrt2}{2}$
i work that out to be $\displaystyle a=\frac{c\sqrt2+d\sqrt2-4\sqrt2}{2}$

Thanks!

2. One point M on the deretrix is (0,1). From this point draw a perpendicular line to the diretrix. Let this line meet the ellipse at P (x,y). Let S be the focus.
According to the definition of the ellipse, SP/PM = e
Find SP and PM and substitute in the above equation and simplify to get the required equation of the ellipse.

3. Hello arze
Originally Posted by arze
Use the locus definition of an ellipse to find the equation of the ellipse with focus (3,4), corresponding directrix line x+y=1 and eccentricity 1/2.
I think the method you're recommended to use is like this.

Suppose that $\displaystyle P (h,k)$ is a general point on the ellipse. Then its distance from the focus $\displaystyle F(3,4)$ is:
$\displaystyle FP = \sqrt{(h-3)^2+(k-4)^2}$
and its distance from the line $\displaystyle x + y -1 = 0$ ($\displaystyle =PD$, say) is:
$\displaystyle PD = \frac{|h+k-1|}{\sqrt{1^2+1^2}}= \frac{|h+k-1|}{\sqrt{2}}$ (See here for this formula.)
So, using the locus definition of the ellipse:
$\displaystyle \frac{FP}{PD}=e=\frac12$

$\displaystyle PD=2FP$

$\displaystyle \Rightarrow |h+k-1|=2\sqrt2\sqrt{(h-3)^2+(k-4)^2}$

$\displaystyle \Rightarrow (h+k-1)^2=8\Big((h-3)^2+(k-4)^2\Big)$

$\displaystyle \Rightarrow h^2+2hk+k^2 - 2h-2k+1=8(h^2-6h+9+k^2-8k+16)$

$\displaystyle \Rightarrow 7h^2-2hk+7k^2-46h-62k+199=0$
So the equation of the locus of $\displaystyle P$ is:
$\displaystyle 7x^2-2xy+7y^2-46x-62y+199=0$