Question

If any tangent drawn to the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) touches one of the circles \(x^2 + y^2 = \alpha^2\), then the range of \(\alpha\) is

• A. \(9 \le \alpha \le 16\)
• B. \(16 \le \alpha \le 25\)
• C. \(3 \le \alpha \le 4\)
• D. \(4 \le \alpha \le 6\)

Hint:

The "pedal curve" properties of an ellipse state that the locus of the foot of the perpendicular from the center to any tangent lies in the annulus bounded by the auxiliary circle (\(r=a\)) and the inscribed circle (\(r=b\)). Thus, the distance \(p\) satisfies \(b \le p \le a\).

Answer 1

The Correct Option is C

Step 1: Understand the Geometry:
We have a family of circles centered at the origin with radius \(\alpha\). A tangent to the ellipse is also a tangent to one of these circles. This means the perpendicular distance from the center (0,0) to the tangent of the ellipse must be equal to the radius of that circle (\(\alpha\)).
Step 2: Range of Perpendicular Distance:
For an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the perpendicular distance \(p\) from the center to any tangent lies between the lengths of the semi-minor axis \(b\) and the semi-major axis \(a\). \[ b \le p \le a \] Here, \(a^2 = 16 \implies a = 4\) and \(b^2 = 9 \implies b = 3\).
Step 3: Determine Range of \(\alpha\):
Since \(\alpha = p\), the range of \(\alpha\) is: \[ 3 \le \alpha \le 4 \]