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 $S'$.
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.
Best Answer
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:
$<1,3>/\sqrt{10}\cdot <1,-1>/\sqrt{2}=-1/\sqrt{5}$
$<1-2\alpha, -2\beta-1>/3\sqrt{10} \cdot <1,-1>/\sqrt{2}=1/\sqrt{5}$
$(1-2\alpha+2\beta+1)=6\implies \beta=\alpha+2$
$(2\alpha-1)^2+(2\alpha+5)^2=90$
$4\alpha^2-4\alpha+1+4\alpha^2+20\alpha+25=90$
$8\alpha^2+16\alpha-64=0$
$\alpha^2+2\alpha+1=9\implies \alpha \in\{2,-4 \}\implies \beta \in \{4,-2\}$
$|\alpha+\beta|=6$