Question:

<p><strong>Step 1: Identify the endpoints of the latus rectum.</strong></p><p>For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the endpoints of the latus rectum are at $x = ae$ where $e = \sqrt{1 - \frac{b^2}{a^2}}$.</p><p>At these points, $y = \pm\frac{b^2}{a}$.</p><p>So the endpoints are: $\left(ae, \frac{b^2}{a}\right)$ and $\left(ae, -\frac{b^2}{a}\right)$ (and their reflections on the left).</p><p><strong>Step 2: Write the tangent equation at the endpoint.</strong></p><p>The tangent to the ellipse at point $(x_1, y_1)$ is: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$</p><p>At the point $\left(ae, \frac{b^2}{a}\right)$:</p><p>$$\frac{x \cdot ae}{a^2} + \frac{y \cdot \frac{b^2}{a}}{b^2} = 1$$</p><p>$$\frac{ex}{a} + \frac{y}{a} = 1$$</p><p><strong>Step 3: Substitute the eccentricity relation.</strong></p><p>Since $e = \sqrt{1 - \frac{b^2}{a^2}}$, we have $e^2 = 1 - \frac{b^2}{a^2}$, so $b^2 = a^2(1-e^2)$.</p><p>The tangent equation becomes:</p><p>$$ex + y = a$$</p><p><strong>Step 4: Find the point independent of b (or e).</strong></p><p>We need to find a point $(0, k)$ on the y-axis that lies on all such tangents for different values of b.</p><p>At $(0, k)$: $e(0) + k = a$, giving $k = a$.</p><p>So the tangent passes through $(0, a)$ for all values of b.</p><p><strong>Step 5: Check the tangent at the lower endpoint.</strong></p><p>At $\left(ae, -\frac{b^2}{a}\right)$, the tangent is:</p><p>$$\frac{ex}{a} - \frac{y}{a} = 1$$</p><p>$$ex - y = a$$</p><p>At $(0, k)$: $0 - k = a$, giving $k = -a$.</p><p>So the tangent passes through $(0, -a)$ for all values of b.</p><p><strong>Step 6: Verify with a specific case.</strong></p><p>For any ellipse with the same major axis 2a but varying b, the tangents at the latus rectum endpoints pass through $(0, a)$ and $(0, -a)$.</p><p>$\therefore$ Answer: a, b</p>