Question 1121318
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With the equation {{{y^2=4ax}}} with a positive, the vertex is at the origin, the parabola opens to the right, and the focus S is (a,0).  So point B is (5a,0).<br>
The circle centered at B touches the parabola in two points, Q and R.  The implication is that the circle is tangent to the parabola at those two points; it does not "pass through" the parabola, thus intersecting the parabola is 4 points.<br>
A generic point on the parabola has coordinates (at^2,2at) where t is a parameter.<br>
Let Q(at^2,2at) be the point in the first quadrant where the circle and parabola are tangent.  By symmetry, point R will be (at^2,-2at).<br>
To have the circle just touch the parabola at Q, we need to have radius BQ of the circle perpendicular to the tangent to the parabola at Q.<br>
The slope of BQ is {{{(0-2at)/(5a-at^2) = (-2at)/(5a-at^2) = (-2t)/(5-t^2)}}}.<br>
The slope of the tangent to the parabola at Q is found by evaluating the derivative at Q.<br>
Implicit differentiation gives us<br>
2yy' = 4a
y' = (4a)/(2y) = 2a/y<br>
So the slope of the tangent to the parabola at Q is (2a)/(2at) = 1/t.<br>
We need to have the product of the slopes equal to -1:<br>
{{{((-2t)/(5-t^2))*(1/t) = -1}}}
{{{(-2)/(5-t^2) = -1}}}
{{{5-t^2 = 2}}}
{{{t^2 = 3}}}
{{{t = sqrt(3)}}}<br>
The coordinates of point Q are (at^2,2at) = (3a,2a*sqrt(3)).<br>
So the coordinates of point R are (at^2,-2at) = (3a,-2a*sqrt(3)).<br>
For the case a=1, B is (5,0) and Q is (3,2*sqrt(3)); the Pythagorean Theorem tells us the radius of the circle is 4.
Below is a graph of the positive branches of the two curves for the case a=1.  The two graphs are {{{y=2*sqrt(x)}}} and {{{(x-5)^2+y^2 = 16}}} which graphs as {{{y = sqrt(16-(x-5)^2)}}}<br>
The intersection of the two graphs is the point Q.<br>
{{{graph(400,400,-2,8,-2,8,2*sqrt(x),sqrt(16-(x-5)^2))}}}