<s>[INST] 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
 [/INST] We have


*

*Sum of distances from focii $=2a = 4\sqrt{10}$ from which we obtain


$$(\alpha-1)^2+(\beta-1)^2 = 90$$


*Product of distances from the tangent (which is helpfully $x=y$) is $b^2$ ($b$ is the semi-minor axis) and hence $$4(\alpha-\beta)=4b^2$$ (Noting that $\alpha>\beta$)

*Distance between foci $=\sqrt{4(a^2-b^2)}$ 
This yields $$\alpha^2+(\beta+2)^2 = 160-4b^2$$
From these we obtain that $3 \alpha-\beta = 34$. Substituting in the first equation we get $\alpha=10$ and hence $\beta = 4$.
Thus $c=5, d=1$ and hence $c+d=6$
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