Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.9 Junior Find the answer

Given a line $l$ with a slope of $\frac{\sqrt{2}}{2}$ that intersects the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at two distinct points, and the projections of these two intersection points on the $x$ -axis are exactly the two foci of the hyperbola. Determine the eccentricity of the hyperbola.
The options are:
A: $\sqrt[4]{2}$
B: $\sqrt{2}$
C: $\sqrt[4]{3}$
D: $\sqrt{3}$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

According to the given information, let $x = \pm c$ and thus $y = \pm \frac{b^2}{a}$ .

Then we have $\frac{\frac{2b^2}{a}}{2c} = \frac{\sqrt{2}}{2}$ , which simplifies to $\frac{b^2}{ac} = \frac{\sqrt{2}}{2}$ .

Now, using $b^2 = c^2 - a^2$ , we have $\frac{c^2 - a^2}{ac} = \frac{\sqrt{2}}{2}$ . This further simplifies to $2c^2 - 2a^2 = \sqrt{2}ac$ .

Dividing both sides by $a^2$ gives us $2(\frac{c^2}{a^2}) - 2 = \sqrt{2}(\frac{c}{a})$ , or $2e^2 - \sqrt{2}e - 2 = 0$ (where $e$ is the eccentricity, given by $\frac{c}{a}$ ).

Solving this quadratic equation, we get $e = \sqrt{2}$ or $e = -\frac{\sqrt{2}}{2}$ (which is discarded since eccentricity cannot be negative).

Therefore, the eccentricity of the hyperbola is $\boxed{e = \sqrt{2}}$ .

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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.