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The locus of foot of the perpendiculars ...

The locus of foot of the perpendiculars drawn from the vertex on a variable tangent to the parabola `y^2 = 4ax` is

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Let foot of the perpendicular from vertex on tangent be M(h,k).
Slope of OM `=(k)/(h)`
`:.` Slope of tangent `=-(h)/(k)`
Thus, equation of tangent is
`y-k=-(h)/(k)(x-h)`
`ory=-(h)/(k)x+(h^(2)+k^(2))/(k)`
Comparing with `y=mx+(a)/(m)`, we have
`m=-(h)/(k)and(a)/(m)=(h^(2)+k^(2))/(k)`
`:." "-(ak)/(h)=(h^(2)+k^(2))/(k)` (eliminating m)
So, `x(x^(2)+y^(2))+ay^(2)=0` is the required locus.
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Knowledge Check

  • If the line lx + my +n=0 is a tangent to the parabola y^(2)= 4ax , then-

    A
    `an^(2)=ml`
    B
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    C
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    A
    `2x-3y+4z=25`
    B
    `2x-3y+4z=20`
    C
    `2x-3y+4z=29`
    D
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