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Question

The locus of point of intersection of two normals drawn to the parabola y2=4ax which are at right angles is

A
y2=a(x3a)
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B
y2=a(xa)
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C
y2=3a(x2a)
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D
y2=2a(x2a)
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Solution

The correct option is A y2=a(x3a)
The given parabola is y2=4ax
Let t denote the parameter, then the equation of the normal to this parabola is given by y=tx+2at+at3
Now this normal passes through the point (h,k) then
k=th+2at+at3at3+t(2ah)k=0 ....... (A)
Let m1,m2,m3 be the roots of this equation.
Let the perpendicular normals correspond to the values of m1,m2
So, m1m2=1
Now from equation (A), m1m2m3=ka
m1m2=1 gives m3=ka
Since m3 is a root of equation (A), we have
a(ka)3ka(2ah)k=0k2=a(h3a)
So locus of (h,k) is y2=a(x3a)

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