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Question

Find the locus of the point of intersection of normals drawn at the extremities of a focal chord of the parabola $$\displaystyle y^{2}=4ax.$$


A
y2=a(x3a)
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B
y2=a(x+3a)
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C
y2=a2(x3a)
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D
y2=a6(x3a)
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Solution

The correct option is A $$\displaystyle y^{2}=a\left ( x-3a \right )$$
Point of intersection of the normals drawn at the ends of a chord of $$y^2=4ax$$ is $$(x,y)=[a(t_{1}^{2}+t_{2}^{2}+t_1t_2+2), -at_1t_2(t_1+t_2)]$$
Since, for focal $$t_1 t_2 =-1$$
$$x=a(t_{ 1 }^{ 2 }+t_{ 2 }^{ 2 }+t_{ 1 }t_{ 2 }+2=a\left[ { \left( { t }_{ 1 }+{ t }_{ 2 } \right)  }^{ 2 }-t_{ 1 }t_{ 2 }+2 \right] =a\left[ { \left( { t }_{ 1 }+{ t }_{ 2 } \right)  }^{ 2 }+3 \right] $$
and $$y=-at_{ 1 }t_{ 2 }(t_{ 1 }+t_{ 2 })=a(t_{ 1 }+t_{ 2 })$$
eliminating $$t_1 + t_2$$ from $$x$$ and $$y$$, we get 
$$\displaystyle y^{2}=a\left ( x-3a \right )$$

Mathematics

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