Question 11. Plz fast
Q.11. 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
(a) x + y = 0
(b) x - y = 0
(c) xy = 0
(d) xy = 1

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 \\ \mathrm{Let} \mathrm{P}\left(\mathrm{h},\mathrm{k}\right) \mathrm{is} \mathrm{point} \mathrm{P} \\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{of} \mathrm{slope} \mathrm{m} \mathrm{to} \mathrm{circle} \mathrm{is} \\ \mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{1+{\mathrm{m}}^2} \\ \mathrm{Tangent} \mathrm{passes} \mathrm{through} \left(\mathrm{h},\mathrm{k}\right) \\ \mathrm{k}=\mathrm{mh}+\mathrm{a}\sqrt{1+{\mathrm{m}}^2} \\ \Rightarrow {\left(\mathrm{k}-\mathrm{mh}\right)}^2={\mathrm{a}}^2\left(1+{\mathrm{m}}^2\right) \\ \Rightarrow {\mathrm{k}}^2+{\mathrm{m}}^2{\mathrm{h}}^2-2\mathrm{mhk}-{\mathrm{a}}^2-{\mathrm{a}}^2{\mathrm{m}}^2=0 \\ \Rightarrow {\mathrm{m}}^2\left({\mathrm{h}}^2-{\mathrm{a}}^2\right)-2\mathrm{khm}+{\mathrm{k}}^2-{\mathrm{a}}^2=0 \\ \mathrm{If} \mathrm{two} \mathrm{roots} \mathrm{are} {\mathrm{m}}_1 \mathrm{and} {\mathrm{m}}_2 \\ \Rightarrow {\mathrm{m}}_1+{\mathrm{m}}_2=\frac{2\mathrm{kh}}{\left({\mathrm{h}}^2-{\mathrm{a}}^2\right)} \\ \mathrm{As} \mathrm{per} \mathrm{question} {\mathrm{m}}_1=\mathrm{tan\alpha } \mathrm{and} {\mathrm{m}}_2=\mathrm{tan\beta } \\ \mathrm{Given} \mathrm{that} \\ \mathrm{cot\alpha }+\mathrm{cot\beta }=0 \\ \Rightarrow \mathrm{cot\alpha }=-\mathrm{cot\beta } \\ \Rightarrow \frac{1}{\mathrm{tan\alpha }}=-\frac{1}{\mathrm{tan\beta }} \\ \Rightarrow \mathrm{tan\alpha }=-\mathrm{tan\beta } \\ \Rightarrow \mathrm{tan\alpha }+\mathrm{tan\beta }=0 \\ \Rightarrow {\mathrm{m}}_1+{\mathrm{m}}_2=0 \\ \Rightarrow \frac{2\mathrm{kh}}{\left({\mathrm{h}}^2-{\mathrm{a}}^2\right)}=0 \\ \Rightarrow \mathrm{hk}=0 \\ \Rightarrow \mathrm{xy}=0 \end{gathered}$$

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