Step 1: Coordinates of P and Q:
Ellipse: \( \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1 \). Auxiliary Circle: \( x^2 + y^2 = 3^2 = 9 \).
Let \( P = (3\cos\theta, 2\sin\theta) \).
Since Q lies on the auxiliary circle and PQ is perpendicular to the major axis (same x-coordinate), \( Q = (3\cos\theta, 3\sin\theta) \).
Step 2: Equations of Normals:
Normal to Ellipse at P:
\[ \frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2 \]
Substitute \( a=3, b=2 \):
\[ \frac{3x}{\cos\theta} - \frac{2y}{\sin\theta} = 9 - 4 = 5 \quad \dots(1) \]
Normal to Circle at Q passes through the origin (0,0):
Slope \( = \frac{3\sin\theta - 0}{3\cos\theta - 0} = \tan\theta \). Equation is \( y = x\tan\theta \), or \( \sin\theta \cdot x - \cos\theta \cdot y = 0 \).
From this, \( \tan\theta = \frac{y}{x} \). Let R be \( (h,k) \), so \( \tan\theta = \frac{k}{h} \).
From a right triangle with sides \( h, k \), we get \( \sin\theta = \frac{k}{\sqrt{h^2+k^2}} \) and \( \cos\theta = \frac{h}{\sqrt{h^2+k^2}} \).
Step 3: Find Locus of R:
Substitute \( \sin\theta \) and \( \cos\theta \) into equation (1) for point R(h,k):
\[ \frac{3h}{h/\sqrt{h^2+k^2}} - \frac{2k}{k/\sqrt{h^2+k^2}} = 5 \]
\[ 3\sqrt{h^2+k^2} - 2\sqrt{h^2+k^2} = 5 \]
\[ \sqrt{h^2+k^2} = 5 \]
Squaring both sides:
\[ h^2 + k^2 = 25 \]
Locus is \( x^2 + y^2 = 25 \).