Finding value of $c+d$

Question

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

Answer 1

We have

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

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

2. 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 $\alpha>\beta$)

3. 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 $c+d=6$

Answer 2

a possible strategy is to consider the following property: "the tangent line always makes equal angles with the generator lines (i.e. lines passing through the foci)"

by this property you can write down a first equation for the coordinates of $S_2(x,y)$

the second equation is given by: $$|AS_1|+|AS_2|=2a=4 \sqrt{10}$$

once you have $S_2$ you can find easily the center as the midpoint between the foci

Answer 3

Another possible strategy is to consider that the center belong to the circle with center in A and radius $2\sqrt{10}$. By this you can find a first equation for c and d. Then you can also find $S_2$ as the opposite point of $S_1$ wro the center. Finally you use the main equation $$|AS_1|+|AS_2|=4\sqrt{10}$$

Answer 4

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 $c+d = \pm 6$.