Question

pls help me to solve this​

Answer 1

Question:-A normal is drawn at a point (x1,y1) of the parabola y^2=16x and this normal makes equal angle with x and y-axes.Then the point (x1,y1) is

Solution:-As we know any point on the parabola will be in the form of $({ - m}^{2} \: \: 2m)$

such that, slope of the normal is m

As the normal makes equal angle with both x and y axis,so,it's slope will be

$\frac{ + }{ }1$

this can be taken out as follows

The normal will make a right Angle triangle,

let the equal angles be x

then,

x+x+90=180

x=90/3=45=y

and we know,

slope =Tan(a)

Tan(a)=1

similarly the angle can also,equal to 135

and ,Tan(135°)=-1

so,slope of the normal can be +1 or -1

putting this value ,we get

the points as(1,-2m) or(-1,2)

We can also,find the equation of the normal

as we know

equation of normal=

$mx - a {m}^{2} - a {m}^{3} \\ {y}^{2} = 4ax \\ here \: 4a = 16 \\ a = 4$

, eqaution will be x-4-4=0 ==>x-8=0

{hope it helps you}