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  • and <img alt="\mathrm{PB}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BPB%7D"

\mathrm{PA} and \mathrm{PB} are tangents drawn to \mathrm{Y}^{2}=4 \mathrm{ax} from arbitrary point \mathrm{P}.If the angle between tangents is \mathrm{\pi / 4},then locus of point \mathrm{p} is

Option: 1

\mathrm{y^{2}=x^{2}+a^{2}+6 a x}


Option: 2

\mathrm{y^{2}=x^{2}-a^{2}+6 a x}


Option: 3

\mathrm{y^{2}=x^{2}+a^{2}-6 a x}


Option: 4

\mathrm{y^{2}=x^{2}-a^{2}-6 a x}


Answers (1)

best_answer

Let \mathrm{P \equiv(h, k)}

Tangent in terms of \mathrm{m} is
\begin{aligned} & y=m x+\frac{a}{m} \\ & \Rightarrow \quad m^{2} x-m y+a=0 . \end{aligned}

If it passes through (h, k), then
\mathrm{m^{2} h-m k+a=0 .}
Since, angle between these tangents is \mathrm{\frac{\pi}{4}},
\mathrm{\tan \frac{\pi}{4}=\frac{\left|m_{1}-m_{2}\right|}{\left|1+m_{1} m_{2}\right|}}
\mathrm{\Rightarrow \left(m_{1}-m_{2}\right)^{2}=\left(1+m_{1} m_{2}\right)^{2}}
\mathrm{\Rightarrow \left(m_{1}-m_{2}\right)^{2}-4 m_{1} m_{2}=\left(1+m_{1} m_{2}\right)^{2}}
\mathrm{\Rightarrow \frac{k^{2}}{h^{2}}-\frac{4 a}{h}=\left(1+\frac{a}{h}\right)^{2}}
Thus required locus is,
\mathrm{y^{2}=4 a x+(a+x)^{2}}.

Posted by

vishal kumar

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