Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Find the answer

4. Let $F$ be the left focus of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ , and $A$ be a point on the ellipse in the first quadrant. Draw a tangent line from point $A$ to the circle $x^{2}+y^{2}=b^{2}$ , with the point of tangency being $P$ . Then $|A F|-|A P|=$

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solutions — 2

Solution 1

4. $a$ .

Let $F(-c, 0), A(a \cos t, b \sin t)\left(t \in\left(0, \frac{\pi}{2}\right)\right)$ .
$\begin{array}{l} \text { Then }|A F|=\sqrt{(a \cos t+c)^{2}+(b \sin t)^{2}} \\ =\sqrt{a^{2} \cos ^{2} t+2 a c \cos t+c^{2}+\left(a^{2}-c^{2}\right) \sin ^{2} t} \\ =\sqrt{a^{2}+2 a c \cos t+c^{2} \cos ^{2} t}=a+c \cos t \\ \text { Also }|A P|=\sqrt{A O^{2}-O P^{2}} \\ =\sqrt{(a \cos t)^{2}+(b \sin t)^{2}-b^{2}} \\ =\sqrt{\left(a^{2}-b^{2}\right) \cos ^{2} t}=c \cos t, \end{array}$
$\text { Therefore }|A F|-|A P|=a \text {. }$

Solution 2

4. a.

Let $F(-c, 0), A(a \cos t, b \sin t)\left(t \in\left(0, \frac{\pi}{2}\right)\right)$ .
Then $|A F|=\sqrt{(a \cos t+c)^{2}+(b \sin t)^{2}}$
$\begin{array}{l} =\sqrt{a^{2} \cos ^{2} t+2 a c \cos t+c^{2}+\left(a^{2}-c^{2}\right) \sin ^{2} t} \\ =\sqrt{a^{2}+2 a c \cos t+c^{2} \cos ^{2} t}=a+c \cos t . \end{array}$
$\begin{array}{l} \text { Also }|A P|=\sqrt{A O^{2}-O P^{2}} \\ =\sqrt{(a \cos t)^{2}+(b \sin t)^{2}-b^{2}} \\ =\sqrt{\left(a^{2}-b^{2}\right) \cos ^{2} t}=c \cos t, \end{array}$

Therefore, $|A F|-|A P|=a$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.