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:

$\displaystyle 7x^2-2xy+7y^2-46x-62y+199=0$

Grandad