Analytic Geometry – Problem 25: Compute A

For the ellipse $C:\frac{x^2}{3}+\frac{y^2}{2}=1$, let $F$ be its right focus. Point $A$ moves on the line $x=3$. From $A$, draw the two tangents $AM$ and $AN$ to the ellipse, touching it at $M$ and $N$. Compute $$|MF|+|NF|-|MN|.$$

A. 3 B. 2 C. 1 D. 0

For $\frac{x^2}{3}+\frac{y^2}{2}=1$, we have $a^2=3$, $b^2=2$, so $c^2=a^2-b^2=1$ and the right focus is $F(1,0)$.

Let $M(x_1,y_1)$ and $N(x_2,y_2)$ be the tangency points. The tangent line to $\frac{x^2}{3}+\frac{y^2}{2}=1$ at $(x_0,y_0)$ is $$\frac{xx_0}{3}+\frac{yy_0}{2}=1.$$ Thus the tangents at $M$ and $N$ are $$\frac{xx_1}{3}+\frac{yy_1}{2}=1,\qquad \frac{xx_2}{3}+\frac{yy_2}{2}=1.$$ Let $A(3,t)$. Since $A$ lies on both tangents, $$x_1+\frac{t y_1}{2}=1,\qquad x_2+\frac{t y_2}{2}=1.$$ So $$\frac{1-x_1}{y_1}=\frac{1-x_2}{y_2}.$$ This means the points $M(x_1,y_1)$, $N(x_2,y_2)$, and $F(1,0)$ are collinear (they have the same slope from $F$). Hence $F$ lies on segment $MN$, so $$|MF|+|NF|=|MN|.$$ Therefore $$|MF|+|NF|-|MN|=0,$$ so the correct choice is D.

$|MF|+|NF|-|MN|=0$ (option D).