Find center of ellipse provided coordinates of one focus, and point on ellipse with it's tangent

Question

At a point $A(1,1)$ on ellipse, equation of tangent is $y=x$. If one of the foci of ellipse is $(0,-2)$ and the coordinates of centre of ellipse are $(\alpha,\beta)$ then find the value of $\alpha+\beta$. It is given that the length of major axis of the ellipse is $4\sqrt{10}$ units.

Solution given

Let $S(0,-2)$ and $S'$ be foci of ellipse. Then the slope of $AS'=\frac13$ and $AS'=3\sqrt{10}$

So the coordinates of $S'$ will be $(10,4)$ and centre is mid point of $S$ and $S'$.

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I understand why the slope of $AS'$ is $\frac13$ but I cannot understand why $AS'$ has to be $3\sqrt{10}$. If I can prove former, I can easily calculate the coordinates of $S'$.

Please help me. Any help is greatly appreciated.

Answer 1

You can use the fact that for all points on the ellipse, the sum of the two distances to the foci is equal to the length of major axis.

So, we have $AS+AS'=4\sqrt{10}$.

Answer 2

Other focus is $(2\alpha, 2(\beta+1))$ since the center of the ellipse is the midpoint of the foci.

The sum of the distances to the foci is constant for an ellipse, twice the semi-major axis$=4\sqrt{10}$ here.

Distance from $(1,1)$ to $(0,-2)$ is $\sqrt{10}$ which means $(2\alpha, 2(\beta+1))$ is $3\sqrt{10}$ from $(1,1)$.

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

Law of reflection:

$<1,3>/\sqrt{10}\cdot <1,-1>/\sqrt{2}=-1/\sqrt{5}$

$<1-2\alpha, -2\beta-1>/3\sqrt{10} \cdot <1,-1>/\sqrt{2}=1/\sqrt{5}$

$(1-2\alpha+2\beta+1)=6\implies \beta=\alpha+2$

$(2\alpha-1)^2+(2\alpha+5)^2=90$

$4\alpha^2-4\alpha+1+4\alpha^2+20\alpha+25=90$

$8\alpha^2+16\alpha-64=0$

$\alpha^2+2\alpha+1=9\implies \alpha \in\{2,-4 \}\implies \beta \in \{4,-2\}$

$|\alpha+\beta|=6$