Answer to Question #120869 in Analytic Geometry for Nikhil

Question

Let y^2=4ax be a parabola and P be a point on it. Let the normal at P intersect the X-axis at Q. Draw a line at Q perpendicular to the above normal. Show that this line intersects the parabola y^2+4a(x-2a)=0

Answer 1

Let the co-ordinates of the point P be $(4a,4a)$ . 

Let Slope of tangent at point p is given by $m_1$.

$y = \sqrt{4ax}$

$m_1 = \frac{dy}{dx} = \frac{\sqrt{4a}}{2{\sqrt{x}}} = \frac{1}{2}$

Let Slope of normal at P, is $m_2$.

$m_2 = -2$

Equation of normal will be

$(y-4a)=-2(x-4a)$

$y = -2x + 12a$

Since this equation will intersect x-axis at Q then co-ordinates of Point Q is $(6a,0)$.

Equation of line passing through point Q and perpendicular to PQ will have slope $\frac{1}{2}.$

$y- 0 = \frac{1}{2}(x - 6a)$

$y = \frac{1}{2}x - 3a$ . . . . . . . . . . .. . . . . . .. .(i)

Equation (i) will intersect parabola $y^{2} + 4a(x-2a) = 0$ at $(-2a,-4a).$

This is all is shown in figure.