Question

PG, the normal at P to a parabola, cuts the axis in G and is produced to Q so that \[GQ = \dfrac{1}{2}PG\]. Prove that the other normals which pass through Q intersect at right angles.

Answer 1

Hint: Assume the equation of parabola as \[{y^2} = 4ax\]. The equation of normal to the parabola at the point \[(a{t^2},2at)\] is given by \[y = - tx + 2at + a{t^3}\]. Use this formula to prove that the other normals which pass through Q intersect at right angles.

Complete step by step answer:

Let us assume that the equation of the parabola is \[{y^2} = 4ax\].

Then the equation of the normal that passes through a point P \[(a{t^2},2at)\] is given as follows:

 \[y = - tx + 2at + a{t^3}............(1)\]

It is given that the normal intersects the axis at G. Hence, the y coordinate of G is zero. Now, we find the x coordinate using equation (1).

\[0 = - tx + 2at + a{t^3}\]

Solving for x, we have:

\[tx = 2at + a{t^3}\]

Dividing both sides with t, we have:

\[x = 2a + a{t^2}\]

Hence, the coordinates of point G is \[(2a + a{t^2},0)\].

It is given that the point G divides the line segment PQ in the ratio 2:1 since PG = 2GQ.

The section formula for a line segment divided in the ratio m:n is given by:

\[(x,y) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\]

Let the coordinates of the point Q be (h, k), then, we have:

\[(2a + a{t^2},0) = \left( {\dfrac{{2h + a{t^2}}}{3},\dfrac{{2k + 2at}}{3}} \right)\]

We now have two equations as follows:

\[2a + a{t^2} = \dfrac{{2h + a{t^2}}}{3}..........(2)\]

\[\dfrac{{2k + 2at}}{3} = 0............(3)\]

From equation (2), we get the value of h.

\[3(2a + a{t^2}) = 2h + a{t^2}\]

Simplifying, we get:

\[6a + 3a{t^2} = 2h + a{t^2}\]

\[6a + 2a{t^2} = 2h\]

Solving for h, we have:

\[h = \dfrac{{6a + 2a{t^2}}}{2}\]

\[h = 3a + a{t^2}\]

From equation (3), solving for the value of k, we get:

\[2k + 2at = 0\]

\[2k = - 2at\]

Dividing both sides by 2, we get:

\[k = - at\]

Hence, the coordinates of the point Q are \[(3a + a{t^2}, - at)\].

From equation (1), we have:

\[a{t^3} + (2a - x)t - y = 0\]

Here, t is the slope of the normal to the parabola.

It is a cubic equation and the product of the roots is given by \[\dfrac{y}{a}\].

\[{t_1}{t_2}{t_3} = \dfrac{y}{a}\]

The equation of normals that pass through Q has the y coordinates as (– at) at Q and one of the tangents has the slope t. Hence, we have:

\[t{t_2}{t_3} = \dfrac{{ - at}}{a}\]

Simplifying, we have:

\[{t_2}{t_3} = - 1\]

Hence, the other two normals that pass through Q intersect at right angles.

Hence, we proved.

Note: You can also similarly assume the parabola \[{x^2} = 4ay\] without any loss of generality and then proceed with the solution by just replacing the x and y coordinates. You will get the same result.