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Prove that the equation of the director circle to the parabola y2=4ax is x + a = 0.

Answer
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Hint: The director circle of a parabola is the directrix of the curve. First draw  the point of intersection of the tangent to the parabola whose equation is y2=4ax. Consider the end points of the focal part on the parabola as P(at2,2at) and P(at12,2at1). Start with differentiating the equation of parabola. 

Complete step-by-step answer:
We know that the equation of parabola is y2=4ax.....(i)

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The figure of the parabola given below illustrates the focus, directrix, axis and vertex of the parabola.
We know that the tangents at the extremities of any focal chord intersect at right angles on the directrix. Let us take the end points of the focal part on the parabola as P(at2,2at) and P(at12,2at1).
To find the equation of tangent to parabola at a given point, we need to differentiate the equation of parabola,
y2=4ax
Differentiating both sides,
2y=4a
dydx=2ay
Therefore, the slope of a tangent at a given point (x1,y1) on the parabola is:
mt=[dydx](x1,y1)=2ay1
Equation of line passing through (x1,y1) having slope m by point – slope form
yy1=m(xx1)
Therefore, equation of tangent
yy1=mt(xx1)
yy1=2ay1(xx1)
y1(yy1)=2ax2ax1
y1yy12=2ax2ax1
y1y4ax1=2ax2ax1
y1y=2ax+2ax1
yy1=2a(x+x1)
At point P(at12,2at), the equation of tangent is
ty=x+at2.....(ii)
At point P(at12,2at1), the equation of tangent is
t1y=x+at12.....(iii)
Therefore, the slope tt1=1
Now, let three tangents intersect at a point (h,k).
The coordinates of points of intersection of tangents in a parabola can be found by subtracting the equation of tangents at the 2 points P and P’ respectively.
i.e. Equation (ii) – Equation (iii)
Subtracting both the equation (a2b2)=(ab)(a+b)
ty=x+at2
t1y=x+at12 
y(tt1)=a(t2t12)
y(tt1)=a(tt1)(t+t1)
y=a(t+t1)
Substitute the value of y in equation (ii).
ty=x+at2
ta(t+t1)=x+at2
at2+att1=x+at2
x=att1
Putting the value of x to find y in equation (ii).
yt=x+at2
yt=att1+at2
yt=at(t1+t)
y=a(t+t1)
(x,y)=(att1,a(t+t1))
(x, y) can be taken as (h, k)
h=att1 and k=a(t+t1).
Since, the tangents are perpendicular, tt1=1
h=a(1)
h=a
h+a=0
Where h = x.
x+a=0
Hence, the locus of the point (h, k) is x = -a which is the equation of the directrix.

Note: The chord joining (at2,2at) and (at12,2at1) passes through focus, if t1t2=1. This is formed because the tangents are perpendicular.