Solving Two Parabola Problems: Proving & Finding Values

[varunKanpur]

I am not able to solve the following problem

#1) Prove that the normal to parabola y²=4ax at (am²,-2am) intersects the parabola again at an angle tan-1(m/2)

What I am thinking is to solve the equation of parabola and equation of normal y=mx-am^-3-2am simultaneously and at that point I will find the slope of tangent and will get the angle between tangent and normal. The problem is that answer is not coming.

#2) For what values of a will the tangents drawn to parabola y²=4ax from a point , not on the y-axis, will be normals to the parabola x²=4y?

I have no idea on how to solve this question
Thanks in Advance

 

[Simon Bridge]

Start with the first one to begin with.

Please show your working?
Show each step with your reasoning.
It can hep to reverse the roles of the x and y axes.

The key to the second one is to rewrite the problem statement in maths.
i.e. a point not on the y-axis is point $p=(p_x,p_y): p_x\neq 0$

 

[varunKanpur]

I got the #1 problem, I was making it more lengthy.

 

[Simon Bridge]

OK - so what about #2?
Same suggestions.

 

[blue_raver22]

I would approach these problems by first understanding the concepts and equations involved. In the first problem, the equation of a parabola is given as y2=4ax, where a is a constant. The point of interest is (am2,-2am), which is on the parabola. The problem asks to prove that the normal to this point intersects the parabola again at an angle of tan-1(m/2).

To solve this, we can use the equation of a normal to a parabola, which is y=mx-am-3-2am. We can substitute the coordinates of the given point into this equation to get the slope of the tangent at that point. From there, we can calculate the angle between the tangent and the normal using trigonometric functions.

In the second problem, we are given two parabolas, y2=4ax and x2=4y, and we need to find the values of a for which the tangents drawn to the first parabola from a point not on the y-axis will be normals to the second parabola. To solve this, we can use the equation of a tangent to a parabola, which is y=mx+a/m. We can then set this equation equal to the equation of the normal to the second parabola, which is y=-1/mx+2/m. By solving for a and setting the discriminant of the resulting quadratic equation equal to 0, we can find the values of a that satisfy this condition.

In both problems, it is important to understand the equations and concepts involved and to use algebra and trigonometry to solve them. It may also be helpful to draw a diagram to better visualize the problem and the relationships between the different elements.