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

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