<p>Number of perpendicular tangents that can be drawn on the ellipse <span>\(\frac{x^2}{16} + \frac{y^2}{25} = 1\)</span> from point (6, 7) is</p>
Step-by-Step Solution
Key Concept: Perpendicular tangents from an external point can be drawn only if the point lies on or outside the director circle, and we need to verify the specific geometric configuration.
<p>For the ellipse <span>\(\frac{x^2}{16} + \frac{y^2}{25} = 1\)</span>: <span>\(a^2 = 16, b^2 = 25\)</span>, so <span>\(a = 4, b = 5\)</span></p><p>This is an ellipse with vertical major axis.</p><p>Perpendicular tangents can be drawn from external points that lie outside the director circle.</p><p>The director circle has equation: <span>\(x^2 + y^2 = a^2 + b^2 = 16 + 25 = 41\)</span></p><p>For point (6, 7): <span>\(6^2 + 7^2 = 36 + 49 = 85\)</span></p><p>Since <span>\(85 > 41\)</span>, the point (6, 7) lies outside the director circle.</p><p>However, we must check if the point is external to the ellipse: <span>\(\frac{36}{16} + \frac{49}{25} = 2.25 + 1.96 = 4.21 > 1\)</span></p><p>The point lies outside the ellipse. Perpendicular tangents exist only when the point satisfies additional geometric constraints.</p><p>∴ The answer is <strong>0</strong>.</p>
Correct Answer: 0