Answer to Question #86616 in Analytic Geometry for RAKESH DEY

Question

find the equation of the normal to the parabola y^2+4x=0 at the point where the line y=x+c touches it

Answer 1

Solution:

First we find the point of intersection of the line

$$y=x+c$$

with the parabola

$$y^2+4x=0:$$

$$x^2+2cx+c^2+4c=0,$$

$$x^2+(2c+4)x+c^2=0,$$

$$D=16(c+1),$$

$$x_1=-c-2-2\sqrt{c+1},x_2=-c-2+2\sqrt{c+1};$$ $$y_1=-2-2\sqrt{c+1},y_2=-2+\sqrt{c+1}$$

So

$$(x_1;y_1)(,x_2;y_2)$$

- wanted points.

Then we have two equations of normal to the parabola at thease points.Find them.

The general equation of normal to the parabola

$$y^2=4ax$$

at the point t is

$$y_0x+2ay-2ay_0-x_0y_0=0$$

Then, when

$$x_1=-c-2-2\sqrt{c+1}$$

and

$$y_1=-2-2\sqrt{c+1}$$

we hawe the equatin of the normal

$$(-2-2\sqrt{c+1})x-2y-6(2+c)-2\sqrt{c+1}(6+c)=0$$

or

$$y=(-1-\sqrt{c+1})x-3(2+c)-\sqrt{c+1}(6+c),$$

when

$$x_2=-c-2+2\sqrt{c+1},$$ $$y_2=-2+2\sqrt{c+1}$$

we hawe the equatin of the normal

$$(-2+2\sqrt{c+1})x-2y-6(2+c)+2\sqrt{c+1}(6+c)=0$$

or

$$y=(-1+\sqrt{c+1})x-3(2+c)+\sqrt{c+1}(6+c).$$

Ansver: equations of the normal to the parabola

$$y^2+4x=0$$

at the point where the line

$$y=x+c$$

touches it is

$$y=(-1-\sqrt{c+1})x-3(2+c)-\sqrt{c+1}(6+c)$$

and

$$y=(-1+\sqrt{c+1})x-3(2+c)+\sqrt{c+1}(6+c).$$