An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.

Question

An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.

Here is a procedure how to do it analiticaly.

• If $T(x_0,y_0)$ is a touching point, then $x_0+y_0=5$
• The equation of ellipse is $${(x_0-{3\over 2})^2\over a^2} +{(y_0+1)^2 \over b^2}=1$$
• Since $2e=1$ we have $a^2-b^2 = {1\over 4}$
• Since the slope of tangent is $-1$ we have $${2(x_0-{3\over 2})\over a^2} -{2(y_0+1)\over b^2}=0$$

And now we have to solve this tedious system. How to do it more geometrical?

Answer 1

The tangency point is that point $P$ on the given line having the minimum sum of distances from foci $A=(1,-1)$ and $B=(2,-1)$. But it is well known how to find such a point: reflect point $B$ about the line, to get $B'=(6,3)$, and $P$ is then the intersection between the given line and line $AB'$. A simple computation gives then $P=(34/9,11/9)$.

Answer 2

The ellipse divides the plane into three regions.

1. the exterior, where the sum of the distances to the foci is greater than $c$.
2. the ellipse itself, where the sum of the distances to the foci is exactly $c$.
3. the interior, where the sum of the distances to the foci is smaller than $c$.

Parametrize the tangent as $x=t,y=5-t$. Thus the sum of the distances to the foci along the tangent has unique global minimum at the tangency point.

Explicitly, the function \begin{align} f(t):={}&\sqrt{(t-1)^2+(5-t+1)^2}+ \sqrt{(t-2)^2+(5-t+1)^2} \\ ={}& \sqrt{(t-1)^2+(6-t)^2}+ \sqrt{(t-2)^2+(6-t)^2} \end{align} has global minimum at $t=\frac {34}9$, yielding the tangency point $(\frac {34}9,\frac{11}9)$. This can be found by differentiating $f(t)$ and imposing $f'(t)=0$, which gives $$ \frac{4 \sqrt{2} (t-4)}{\sqrt{(t-8) t+20}}=-\frac{8 t-28}{\sqrt{2 (t-7) t+37}} $$ which can be squared to a quadratic equation $$ 9 t^2-88 t+204=0\Rightarrow (t-6 ) ( 9 t-34)=0, $$ with roots $6,\frac {34}9$, the latter being the actual solution of the original equation before squaring.

Answer 3

We have the coordinates of any point on the ellipse centered at the origin $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in parametric form as $P(a\cos\theta,b\sin\theta)$, where $\theta$ is a parameter. Now, by using basic calculus, we get the equation of tangent to the ellipse at this point as $\frac{x}{a^2}a\cos\theta+\frac{y}{b^2}b\sin\theta=1$.

Now, by the concept of shifting of origin, we can obtain similar results for the ellipse given in the question.

$\Rightarrow$ Equation of ellipse: $$\frac{(x-\frac32)^2}{a^2}+\frac{(y+1)^2}{b^2}=1$$

$\Rightarrow$ Coordinates of any point on the ellipse in parametric form: $Q(a\cos\theta+\frac32,b\sin\theta-1)$

$\Rightarrow$ Equation of tangent to the ellipse at this point: $$\frac{(x-\frac32)}{a^2}a\cos\theta+\frac{(y+1)}{b^2}b\sin\theta=1$$ $$\Rightarrow \frac xa \cos\theta+\frac yb \sin\theta=\frac{3\cos\theta}{2a}-\frac{\sin\theta}{b}+1$$

Comparing this with the given equation of tangent $x+y=5$, we get $$\frac{\cos\theta}{a}=\frac{\sin\theta}{b}=\frac{\frac{3\cos\theta}{2a}-\frac{\sin\theta}{b}+1}{5}\ldots(1)$$

Also, we have $$a^2-b^2=\frac14\ldots(2)$$

Now, we can assume $\frac{\cos\theta}{a}$ and $\frac{\sin\theta}{b}$ as two single variables and solve for them to get $$\Rightarrow\frac{\cos\theta}{a}=\frac{\sin\theta}{b}=\frac29\ldots(3)$$

$(2)$ and $(3)$ can easily be solved and we get $\cos\theta=\sqrt\frac{41}{81}$ and $ \sin\theta=\sqrt\frac{40}{81}$.

Now you can easily obtain the coordinates of $Q$ by substituting $a$ and $b$ in terms of $\cos\theta$ and $\sin\theta$.

Therefore, $$\Rightarrow Q=(\frac{34}{9},\frac{11}{9})$$