Finding value of $c+d$ 
At a point $A(1,1)$ on the ellipse , equation of tangent is $y=x.$ If one of the
foci of the ellipse is $(0,-2)$ and coordinate of center of ellipse is $(c,d)$.
Then find value of $c+d$ (given length of major axis is $4\sqrt{10} unit$)

Attempt : assuming one foci is at $S_{1}(0,-2)$ and other is at $S_{2}(\alpha,\beta)$ and $A(1,1)$ be a point on ellipse. then $AS_{1}+AS_{2} = 4\sqrt{2}$
$\sqrt{10}+\sqrt{(1-\alpha)^2+(1-\beta)^2} = 4\sqrt{2}$
could some help me to solve it , thanks

The slope of the normal at $(1,1)$ is $-1$ and the normal bisects the angle between $S_1A, S_2A$. Since the slope of $AS_1$ is 3, slope of normal is $-1$, slope of $AS_2$ is $\frac{1}{3}$. The equation of $AS_2$ is $y-1 = \frac{1}{3}(x-1)$. Thus $\beta -1 = \frac{1}{3}(\alpha-1)$. Substitute this in the equation $AS_1+AS_2 = 4\sqrt{10}$ to get $|1-\alpha| = 9$ giving $\alpha = 10, -8$. We can now obtain $\beta$. There are two possible ellipses. When $\alpha = -8$, we get $\beta = -2$ and the center of the ellipse is $(-4, -2)$ and $c+d = -6$. When $\alpha = 10$, $\beta = 4$ and center is $(5,1)$ and $c+d = 6$. Thus $c+d = \pm 6$.
