Question:
The tangents to \(x^2 + y^2 = a^2\) having inclinations \(\alpha\) and \(\beta\) intersect at \(P\). If \(\cot \alpha + \cot \beta = 0\), then the locus of \(P\) is
The tangents to \(x^2 + y^2 = a^2\) having inclinations \(\alpha\) and \(\beta\) intersect at \(P\). If \(\cot \alpha + \cot \beta = 0\), then the locus of \(P\) is
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Use sum of roots of quadratic in \(m\).
Updated On: Apr 23, 2026
- \(x + y = 0\)
- \(x - y = 0\)
- \(xy = 0\)
- None of these
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The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ y = mx + a\sqrt{1+m^2} \text{ is tangent to } x^2+y^2=a^2 \]
Step 2: Calculation / Simplification}
Let \(P(h, k)\). Tangent passes through \(P\): \(k = mh + a\sqrt{1+m^2}\)
\((k - mh)^2 = a^2(1+m^2)\)
\(m^2(h^2 - a^2) - 2hkm + (k^2 - a^2) = 0\)
Roots \(m_1 = \tan \alpha, m_2 = \tan \beta\)
\(m_1 + m_2 = \frac{2hk}{h^2 - a^2}\)
\(\cot \alpha + \cot \beta = \frac{1}{m_1} + \frac{1}{m_2} = \frac{m_1+m_2}{m_1m_2} = 0 \Rightarrow m_1+m_2 = 0\)
\(\frac{2hk}{h^2 - a^2} = 0 \Rightarrow hk = 0\)
Locus: \(xy = 0\)
Step 3: Final Answer
\[ xy = 0 \]
\[ y = mx + a\sqrt{1+m^2} \text{ is tangent to } x^2+y^2=a^2 \]
Step 2: Calculation / Simplification}
Let \(P(h, k)\). Tangent passes through \(P\): \(k = mh + a\sqrt{1+m^2}\)
\((k - mh)^2 = a^2(1+m^2)\)
\(m^2(h^2 - a^2) - 2hkm + (k^2 - a^2) = 0\)
Roots \(m_1 = \tan \alpha, m_2 = \tan \beta\)
\(m_1 + m_2 = \frac{2hk}{h^2 - a^2}\)
\(\cot \alpha + \cot \beta = \frac{1}{m_1} + \frac{1}{m_2} = \frac{m_1+m_2}{m_1m_2} = 0 \Rightarrow m_1+m_2 = 0\)
\(\frac{2hk}{h^2 - a^2} = 0 \Rightarrow hk = 0\)
Locus: \(xy = 0\)
Step 3: Final Answer
\[ xy = 0 \]
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