<p>The locus of point of intersection of perpendicular tangents of ellipse \(\frac{(x-1)^2}{16} + \frac{(y-1)^2}{9} = 1\) is:</p>
<p>(a) \(x^2 + y^2 = 25\)</p>
<p>(b) \(x^2 + y^2 + 2x + 2y - 23 = 0\)</p>
<p>(c) \(x^2 + y^2 - 2x - 2y - 23 = 0\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: The locus of intersection points of perpendicular tangents to an ellipse is its director circle, given by \((x-h)^2 + (y-k)^2 = a^2 + b^2\) for an ellipse centered at \((h,k)\).
<p>For an ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the director circle (locus of intersection of perpendicular tangents) is given by: \((x-h)^2 + (y-k)^2 = a^2 + b^2\)</p><p>Here \(h=1, k=1, a^2=16, b^2=9\)</p><p>So: \((x-1)^2 + (y-1)^2 = 16 + 9 = 25\)</p><p>Expanding: \(x^2 - 2x + 1 + y^2 - 2y + 1 = 25\)</p><p>\(x^2 + y^2 - 2x - 2y + 2 = 25\)</p><p>\(x^2 + y^2 - 2x - 2y - 23 = 0\)</p><p>∴ Answer is (c).</p>
Correct Answer: c