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

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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