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The tangents to \mathrm{x^{2}+y^{2}=a^{2}} having inclinations \mathrm{\alpha} and \mathrm{\beta } intersect at \mathrm{P }. If \mathrm{\cot \alpha+\cot \beta=0 },then the locus of \mathrm{P }is

Option: 1

\mathrm{x+y=0}


Option: 2

\mathrm{x-y=0}


Option: 3

\mathrm{x y=0}


Option: 4

None of these.


Answers (1)

best_answer

Let the coordinates of \mathrm{P} be \mathrm{(h, k)}. Let the equation of a tangent from \mathrm{P(h, k)} to the circle \mathrm{x^{2}+y^{2}=a^{2}}be

\mathrm{y=m x+a \sqrt{1+m^{2}}}

Since, \mathrm{P(h, k)} lies on \mathrm{y=m x+a \sqrt{1+m^{2}}}, therefore,

\mathrm{k}=\mathrm{mh}+\mathrm{a} \sqrt{1+\mathrm{m}^{2}} \text { P }(k-m h)^{2}=a \sqrt{(1+m)^{2}}
\text { p } \mathrm{m}^{2}\left(\mathrm{k}^{2}-\mathrm{a}^{2}\right)-2 m k h+\mathrm{h}^{2}-\mathrm{a}^{2}=0

The is a quadratic in \mathrm{m}.Let the two roots be \mathrm{m_{1}} and \mathrm{m_{2}}, then

\mathrm{m_{1}+m_{2}=\frac{2 h k}{k^{2}-a^{2}}}
But \mathrm{\tan \alpha=m_{1}, \tan \beta=m_{2}} and it s given that

\mathrm{\cot \alpha+\cot \beta=0}.

\mathrm{\frac{1}{m_{1}}+\frac{1}{m_{2}}=0\: p \: m_{1}+m_{2}=0 \quad p \: \frac{2 h k}{k^{2}-a^{2}}=0}

\mathrm{\Rightarrow \quad h k=0}. Hence, the locus of \mathrm{(h, k)} is \mathrm{x y=0}.

Hence (C) is the correct answer.

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