Ellipse
Foci trajectory
Grade 11
Question:
<p>An ellipse has semi-major axis of length 2 and semi-minor axis of length 1. It slides between the co-ordinate axes in the first quadrant, while maintaining contact with both x-axis and y-axis. The locus of the foci of the ellipse is:</p>
<p>(a) \(x^2 + y^2 + \frac{1}{x^2} + \frac{1}{y^2} = 16\)</p>
<p>(b) \(x^2 + y^2 + \frac{1}{x^2} - \frac{1}{y^2} = 2\sqrt{3} + 4\)</p>
<p>(c) \(x^2 + y^2 - \frac{1}{x^2} - \frac{1}{y^2} = 2\sqrt{3} + 4\)</p>
<p>(d) \(x^2 - y^2 + \frac{1}{x^2} - \frac{1}{y^2} = 2\sqrt{3} + 4\)</p>
Step-by-Step Solution
Key Concept: The foci of an ellipse are at a fixed distance from the centre; as the centre moves along the locus, the foci trace their own curve.
<p>For the ellipse with semi-major axis \(a = 2\) and semi-minor axis \(b = 1\), the distance from centre to focus is \(c = \sqrt{a^2 - b^2} = \sqrt{4 - 1} = \sqrt{3}\). The foci are located at distance \(\sqrt{3}\) from the centre along the major axis. As the ellipse slides maintaining contact with both axes with centre at (h, k) where \(h^2 + k^2 = 5\), the foci trace a path. The locus equation is \(x^2 + y^2 + \frac{1}{x^2} + \frac{1}{y^2} = 16\).</p>
Correct Answer: A