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  • Find the locus of the point of intersection of normals at points <img alt="\mathrm{A\left(t_1\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%5Cleft%28t_1%5Cright%29%7D"

Find the locus of the point of intersection of normals at points \mathrm{A\left(t_1\right)} and \mathrm{B\left(t_2\right)} on the parabola \mathrm{\mathrm{y}^2=4 \mathrm{ax}}, such that x-axis cuts the chord \mathrm{\mathrm{AB}} in the ratio \mathrm{2: 1.}

Option: 1

\mathrm{4 y^2=\frac{29}{3} a x}


Option: 2

\mathrm{7 y^2=4 a x}


Option: 3

\mathrm{7 y^2=a x}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

\mathrm{A C: C B=2: 1} where C lie on x-axis

\mathrm{ \Rightarrow \frac{4 \mathrm{at}_2+2 \mathrm{at}_1}{3}=0 \Rightarrow \mathrm{t}_1=-2 \mathrm{t}_2 }

Hence coordinates of T are \mathrm{T \equiv\left(a t_1 t_2, a\left(t_1+t_2\right)\right)}

\mathrm{ \mathrm{T} \equiv\left(-2 \mathrm{at}_2^2,-\mathrm{at}{ }_2\right) \text { and } \mathrm{A} \equiv\left(4 \mathrm{at}_2^2,-4 \mathrm{at}_2\right) }

Since \mathrm{\angle P A T=\angle P B T=90^{\circ}}

\mathrm{\Rightarrow} PATB is concylic and PT is one of the diameter

\mathrm{ \begin{aligned} & \Rightarrow h-2 a t_2^2=4 a t_2^2+a t_2^2 \Rightarrow h=7 a t_2^2 \\ & \Rightarrow t_2^2=\frac{h}{7 a} \end{aligned} }
and \mathrm{\mathrm{k}-\mathrm{at}_2=-2 \mathrm{at}_2 \Rightarrow 7 \mathrm{k}^2=\mathrm{ah}}

Hence the locus is \mathrm{7 y^2=a x.}

Posted by

Gaurav

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