The locus of the foot of the perpendicular dr | vedclass

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(A) The equation of any tangent to the parabola $y^2 = 4ax$ is given by $y = mx + \frac{a}{m}$,where $m$ is the slope. The line perpendicular to this

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$x = 0$

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(A) The equation of any tangent to the parabola $y^2 = 4ax$ is given by $y = mx + \frac{a}{m}$,where $m$ is the slope. The line perpendicular to this tangent passing through the focus $(a, 0)$ has a slope of $-\frac{1}{m}$. The equation of this perpendicular line is $y - 0 = -\frac{1}{m}(x - a)$,which simplifies to $y = -\frac{x}{m} + \frac{a}{m}$. To find the locus of the foot of the perpendicular,we eliminate the parameter $m$ from the two equations: $1) \ y = mx + \frac{a}{m} \implies my = m^2x + a$ $2) \ y = -\frac{x}{m} + \frac{a}{m} \implies my = -x + a$ Equating the two expressions for $my$: $m^2x + a = -x + a$ $m^2x = -x$ $x(m^2 + 1) = 0$ Since $m^2 + 1 
eq 0$ for real $m$,we must have $x = 0$. Thus,the locus is the tangent at the vertex,which is the $y$-axis,represented by $x = 0$.

The locus of the foot of the perpendicular drawn from the focus to any tangent of the parabola $y^2 = 4ax$ is:

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