# Angles in an Ellipse proof

#### clem678ens

Hello,

"Let A be a point on an ellipse with the focal point B1 near vertex S1 and the focal point B2 at vertex S2. Let C be the intersection of the perpendicular from B2 to the tangent through point A. Prove: The angles ACS1 and B2CS2 are equal."

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#### skipjack

Forum Staff
Let A be the point (a*cos(t), b*sin(t)), where a > b > 0, c = √(a² - b²) and π/2 < t < π.
Let S$_1$ and S$_2$ be the points (-a, 0) and (a, 0) respectively.
Let the ellipse have equation (x/a)² + (y/b)² = 1.
Its tangent at A has equation (cos(t)/a)x + (sin(t)/b)y = 1 and slope -(b/a)cot(t).
The slope of any perpendicular to it is (a/b)tan(t).
Hence the perpendicular through B$_2$(c, 0) has equation y = (a/b)tan(t)(x - c).
Find the coordinates of C and prove C lies on the circle with equation x² + y² = a².

topsquark

#### Walagaster

I think this is an interesting problem. I have redrawn it in the image below. You start with the ellipse and its two foci $f_1,~f_2$ and a line tangent at $T$. Then draw a perpendicular from $f_2$ to the tangent line intersecting it at $P$. The two angles you are asked to show equal are in red and blue. The two corresponding sides $TP$ and $PF_2$of the two angles are perpendicular by construction. The other sides $AP$ and $PB$ will be perpendicular when $P$ is on the circumscribing circle with diameter $AB$. So, to rephrase the problem: Show that a perpendicular line from a focus to a tangent line intersects it on the circumcircle of the ellipse. It seems like there should be a purely geometric argument for that. Could be well known for all I know. Anybody?

#### skipjack

Forum Staff
The property is known (it can be found by a web search).