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

 Tangents \mathrm{PA} and \mathrm{PB} are drawn to parabola \mathrm{y}^{2}=4 \mathrm{ax}. If slope of bisector of the angle \mathrm{P A B} is \mathrm{\sqrt{3}}, then locus of point \mathrm{'p'} is

Option: 1

\mathrm{y^{2}=x^{2}+3 a^{2}-2 a x}


Option: 2

\mathrm{y^{2}=3 x^{2}+3 a^{2}-2 a x}


Option: 3

\mathrm{y}^{2}=\mathrm{x}^{2}+3 \mathrm{a}^{2}+10 \mathrm{ax}


Option: 4

\mathrm{y}^{2}=3 \mathrm{x}^{2}+3 \mathrm{a}^{2}+10 \mathrm{ax}


Answers (1)

best_answer

Slope of tangents will be \tan \left(\frac{\pi}{3}+\theta\right), \tan \left(\frac{\pi}{3}-\theta\right),

Let \mathrm{A \equiv\left(a^{t_{1}^{2}}, 2 a^{t_{1}}\right), B \equiv\left(a^{t_{2}^{2}}, 2 a t_{2}\right)}
\mathrm{\Rightarrow \quad P \equiv\left[a t_{1} t_{2}, a\left(t_{1}+t_{2}\right)\right]}

We have
\mathrm{\mathrm{t}_{1}=\cot \left(\frac{\pi}{3}+\theta\right), \mathrm{t}_{2}=\cot \left(\frac{\pi}{3}-\theta\right)},
\mathrm{\Rightarrow \mathrm{t}_{1}=\frac{\frac{1}{\sqrt{3}} \cdot \cot \theta-1}{\frac{1}{\sqrt{3}}+\cot \theta}, \mathrm{t}_{2}=\frac{\frac{1}{\sqrt{3}} \cdot \cot \theta+1}{\frac{1}{\sqrt{3}}-\cot \theta} }
\mathrm{\Rightarrow \quad \cot \theta=\frac{\sqrt{3}+\mathrm{t}_{1}}{1-\sqrt{3} \mathrm{t}_{1}}=\frac{\mathrm{t}_{2}-\sqrt{3}}{1+\sqrt{3} \mathrm{t}_{2}} }
\mathrm{\Rightarrow \quad \sqrt{3}+3 \mathrm{t}_{2}+\mathrm{t}_{1}+\sqrt{3} \mathrm{t}_{1} \mathrm{t}_{2}=\mathrm{t}_{2}-\sqrt{3}-\sqrt{3} \mathrm{t}_{1} \mathrm{t}_{2}+3 \mathrm{t}_{1} }
\mathrm{\Rightarrow \quad 2 \sqrt{3}+2 \sqrt{3} \mathrm{t}_{1} \mathrm{t}_{2}=2\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right) }
\mathrm{\Rightarrow \quad 3\left(1+\mathrm{t}_{2} \mathrm{t}_{2}\right)^{2}=\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)^{2}=\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}-4 \mathrm{t}_{1} \mathrm{t}_{2}}
Thus locus of p is
\mathrm{y^{2}=3 x^{2}+3 a^{2}+10 a x}

Posted by

shivangi.shekhar

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