Answer to Question #123536 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 intersects the x-axis at Q. Draw a line at Q perpendicular to the above normal. Show that the line intersects the parabola y^2+4a(x-2a)=0

Answer 1

let us consider the point P(a,2a) on the parabola $y^2=4ax.$

The rough diagram of the question is attached here...

from P the normal to the parabola $y^2=4ax$ is drawn which intersects the x axis at the point Q,the coordinate of which is (3a,0). ( working of Q is shown below).

from Q a line perpendicular to PQ is drawn and we have to show that it intersects the parabola

$y^2+4a(x-2a)=0.$

Now equation of tangent at the point P(a,2a) is given by

$y.2a=2a(x+a)\\or \space y=x+a \space \space ..............(1)$ (Here slope (m) of tangent is 1.)

$\therefore$ slope of the Normal is $-1.$

Now Equation of Normal (which is perpendicular to the tangent at P ) is given by

$(y-2a)=-1(x-a)\\or \space y=-x+3a.\\$

AS this normal intersects X-axis at Q. so ordinate of Q must be '0'.

putting y=0 in $y=-x+3a.$

we get x=3a.

$\therefore$ The coordinate of Q is (3a,0).

Now slope of the line which is passing through Q (3a,0) and perpendicular to the normal at P

is again $1$ as product of slope of two perpendiculars line is always $-1$ .

$\therefore$ The equation of line passing through Q(3a,0) is given by

$(y-0)=1(x-3a)\\or \space y=x-3a.$

Now the coordinates of the end points of the latus rectum of the parabola $y^2=4ax \space is \\$

$P(a,2a) \space and (a,-2a).$

clearly the Coordinate (a,-2a) satisfies both $y=x-3a \space and \space y^2+4a(x-2a)=0.$

Hence the required line $(y=x-3a)$ intersects the parabola $y^2+4a(x-2a)=0.$

(PROVED).