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  • Find the locus of the point of intersection of two normals to the parabola x2 = 4ay, which are at right angles to one anotherOption: 1 <im

Find the locus of the point of intersection of two normals to the parabola x2 = 4ay, which are at right angles to one another

Option: 1

y^{2}=a(x+3 a)


Option: 2

y^{2}=a(x-3 a)


Option: 3

x^{2}=a(y+3 a)


Option: 4

x^{2}=a(y-3 a)


Answers (1)

best_answer

 

 

Point of Intersection of Normal of a Parabola -

Point of Intersection of Normal of a Parabola

Let the equation of parabola be y2 = 4ax 

\\ {\text {Two points, } P \equiv\left(a t_{1}^{2}, 2 a t_{1}\right) \text { and } Q \equiv\left(a t_{2}^{2}, 2 a t_2\right) \text { on the parabola } y^{2}=4 a x .} \\ \\ {\text { Then, equation of Normal; at } P \text { and } Q \text { are }} \\\\ {y_1=-t_{1} x+2 a t_{1}+a t_{1}^{3}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i)\\\\ {y_2=-t_{2} x+2 a t_{2}+a t_{2}^{3}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)\\\\ {\text { solving (i) and (ii) }} \text { we get, }\\\ \\ x=2 a+a\left(t_{1}^{2}+t_{2}^{2}+t_{1} t_{2}\right), y=-a t_{1} t_{2}\left(t_{1}+t_{2}\right)\\\\\text{If R is the point of intersection of two normal then,}\\\\\mathrm{R\equiv \left [ 2a+a(t_1^2+t_2^2+t_1t_2),\;-at_1t_2(t_1+t_2) \right ]}

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The equations of normals at A(t1) and B(t2) are

\begin{array}{l}{x=-t_{1}y+2 a t_{1}+a t_{1}^{3}} \\ {x=-t_{2} y+2 a t_{2}+a t_{2}^{3}}\end{array}

Since these two normals are at right angles, so t1t2 = –1.

\begin{array}{l}{\text { Let P}(h, k) \text { be the point of intersection of two normals. }} \\ {\text { Then, } k=2 \mathrm{a}+a\left(t_{1}^{2}-t_{1} t_{2}+t_{2}^{2}\right)} \\ {\text { and } h=-a t_{1} t_{2}\left(t_{1}+t_{2}\right)} \\ {\Rightarrow \quad k=2 a+a\left\{\left(t_{1}+t_{2}\right)^{2}-2 t_{1} t_{2}\right\}} \\ {\text { and } h=-a t_{1} t_{2}\left(t_{1}+t_{2}\right)} \\ {\Rightarrow \quad k=2 a+a\left\{\left(t_{1}+t_{2}\right)^{2}-1\right\}} \\ {\text { and } \quad h=a\left(t_{1}+t_{2}\right)}\end{array}

Eliminating t1 and t2, we get,

h^{2}=a(k-3 a)

Hence, the locus of \text{P}(h, k) \text { is } x^{2}=a(y-3 a).

Posted by

seema garhwal

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