<p>An ellipse slides between two lines at right angles to one another. Then, the locus of its centre is</p>
Step-by-Step Solution
Key Concept: When an ellipse slides between two perpendicular tangent lines, the locus of its centre satisfies the constraint that the sum of squares of the coordinates remains constant, which is the equation of a circle.
<p><strong>Solution:</strong></p><p>Let the two perpendicular lines be taken as the X-axis and Y-axis. Let $C(a, b)$ be the centre of the ellipse with semi-major axis $a$ and semi-minor axis $b$.</p><p>Let $S(x_1, y_1)$ and $S'(x_2, y_2)$ be the foci of the ellipse.</p><p>Since the ellipse slides between the two perpendicular lines (the axes), they are tangent to the ellipse. From the tangency conditions:</p><p>$x_1 + x_2 = 2a$ ... (i)</p><p>$y_1 + y_2 = 2b$ ... (ii)</p><p>$y_1 y_2 = b^2$ ... (iii)</p><p>$x_1 x_2 = b^2$ ... (iv)</p><p>Also, $S'S^2 = 4a^2e^2$, which gives:</p><p>$(x_1 - x_2)^2 + (y_1 - y_2)^2 = 4a^2e^2$ ... (v)</p><p>Expanding: $\{(x_1 + x_2)^2 + (y_1 + y_2)^2\} - \{(x_1 - x_2)^2 + (y_1 - y_2)^2\} = 4(x_1x_2 + y_1y_2)$</p><p>Substituting: $4a^2 + 4b^2 - 4a^2e^2 = 4(b^2 + b^2) = 8b^2$</p><p>$a^2 + b^2 - a^2e^2 = 2b^2$</p><p>Since $e^2 = 1 - \frac{b^2}{a^2}$, we have $a^2e^2 = a^2 - b^2$</p><p>$a^2 + b^2 - (a^2 - b^2) = 2b^2$</p><p>$2b^2 = 2b^2$ ✓</p><p>The relation gives: $a^2 + b^2 = a^2 + b^2$, which simplifies to $x^2 + y^2 = \text{constant}$ for the locus.</p><p>Hence, the locus of $(a, b)$ is <strong>a circle</strong>.</p>
Correct Answer: A