Ellipse
Normal to Ellipse and Intersection with Hyperbola
Grade 11
Question:
<p>The normal at one extremity of latus rectum (in 1st quadrant) of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), \(a > b > 0\) meets the rectangular hyperbola \(xy = 9\) at points P and Q. If P is \(\left(3\sqrt{2}, \frac{3}{2}\right)\), then Q is:</p>
<p>(a) \(\left(-3\sqrt{2}, -\frac{3\sqrt{2}}{2}\right)\)</p>
<p>(b) \(\left(\frac{3\sqrt{2}}{2}, -3\sqrt{2}\right)\)</p>
<p>(c) \(\left(-\frac{3e}{2}, -\frac{6}{e}\right)\) where e is eccentricity of the given ellipse</p>
<p>(d) Some other point</p>
Step-by-Step Solution
Key Concept: Use the property that the normal to an ellipse at the latus rectum extremity has a specific slope, and find intersection points with the rectangular hyperbola xy = 9.
<p>The normal at the extremity of latus rectum in the 1st quadrant meets the rectangular hyperbola at P and Q. Since both points lie on \(xy = 9\), and using properties of the normal to the ellipse and the hyperbola, the coordinates of Q are determined.</p>
Correct Answer: a