Questions: Consider the vertical line l that goes through the point (x, 0). If -a ≤ x ≤ a, then our line intersects the ellipse. What is the length of the interval of l that is contained inside the ellipse (interval P Q in the figure below)? Note that we expect the answer to depend on a, b and x.

Solution Steps

The equation of an ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$ is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The vertical line $l$ goes through the point $(x, 0)$. We need to find the points where this line intersects the ellipse. Since the line is vertical, its equation is $x = constant$. Substituting $x$ into the ellipse equation, we get: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2}$ $y^2 = b^2(1 - \frac{x^2}{a^2})$ $y^2 = b^2(\frac{a^2 - x^2}{a^2})$ $y = \pm \frac{b}{a}\sqrt{a^2 - x^2}$

So, the coordinates of point P are $(x, \frac{b}{a}\sqrt{a^2 - x^2})$ and the coordinates of point Q are $(x, -\frac{b}{a}\sqrt{a^2 - x^2})$.

The length of the interval PQ is the difference between the y-coordinates of P and Q:

$PQ = \frac{b}{a}\sqrt{a^2 - x^2} - (-\frac{b}{a}\sqrt{a^2 - x^2}) $ $PQ = 2\frac{b}{a}\sqrt{a^2 - x^2}$

Final Answer: