Question

If the tangent to ellipse \(x^2 + 2y^2 = 1\) at point \(P\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)\) meets the auxiliary circle at the points \(R\) and \(Q\), then tangents to circle at \(Q\) and \(R\) intersect at

• A. \(\left(\frac{1}{\sqrt{2}}, 1\right)\)
• B. \(\left(1, \frac{1}{\sqrt{2}}\right)\)
• C. \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
• D. \(\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)\)

Hint:

Intersection of tangents at ends of chord is pole of the chord.

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \text{Tangent at } (x_1, y_1): xx_1 + 2yy_1 = 1 \]
Step 2: Calculation / Simplification}
At \(P(1/\sqrt{2}, 1/2)\): \(\frac{x}{\sqrt{2}} + y = 1 \Rightarrow x + \sqrt{2}y = \sqrt{2}\)
Auxiliary circle: \(x^2 + y^2 = 1\)
Intersection of tangent with circle is chord of contact from \(T(h, k)\)
Chord of contact: \(hx + ky = 1\)
Comparing: \(\frac{h}{1} = \frac{k}{\sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow h = \frac{1}{\sqrt{2}}, k = 1\)
Intersection point \(T(1/\sqrt{2}, 1)\)
Step 3: Final Answer
\[ \left(\frac{1}{\sqrt{2}}, 1\right) \]