Two Points On A Parabola

Calculus Level 3

Two points are chosen at a parabola, such that tangents at these points are perpendicular to each other . Let the angle subtended by the chord joining the points at the vertex of the parabola be $\theta$ . Find the value of $\theta_{max} + \theta_{min}$ in degrees .

The answer is 216.87.

2 solutions

Anish Puthuraya
Mar 29, 2014

As a more convenient approach, we shall consider the parabola $y^2= 4ax$

alt text

It is a known fact that the tangents at the end points of the focal chord (the chord of the parabola which passes through its focus) intersect at $90^o$ to each other on the directrix of the parabola.

Hence, the two points chosen, must lie on the focal chord.

Also, the minimum angle subtended at the vertex by the end points of a focal chord is $90^o$

To find the maximum angle subtended, we consider the focal chord which is drawn perpendicular to the axis of the parabola.

Therefore, the coordinates of the end points of the focal chord become : $\displaystyle (a,2a);(a,-2a)$

The angle subtended by these two points at the vertex is : $\displaystyle \theta_{max} = 2\tan^{-1}2$

$\theta_{max} = 126.87^o$

We already know that $\displaystyle\theta_{min} = 90^o$

Hence,

$\theta_{max} + \theta_{min} = 126.87^o+90^o = \boxed{216.87^o}$

Sorry for the less detailed solution, I didn't have much time to write this..(its almost 1 A.M. in the morning and my mother is scolding me for sitting up late!)

Anish Puthuraya - 7 years, 2 months ago

I would like to know what's wrong with my approach.

I will work in parametric coordinates. The two points are clearly $(at^2,2at)$ and $(a/t^2,-2a/t)$ . ( I am working with the parabola $y^2=4ax$ ).

The angle between the lines made by joining the vertex to the above two points is given by:

$\displaystyle \tan\theta=\left|\frac{2/t+2t}{1-4}\right|=\frac{2}{3}\left|t+\frac{1}{t}\right|$

Finding the maximum and minimum value of the above expression doesn't give me the right answer. Where did I go wrong? :(

Pranav Arora - 7 years, 2 months ago

Note that by using the formula which you used you can only obtain the acute angle between two lines. A better approach would be to use vectors in this case :)

Karthik Kannan - 7 years, 2 months ago

Well, according to your approach, $\tan\theta$ can never be negative. Why should that be the case? Since, $\tan\theta$ being negative implies that $\theta>90^o$ , which I think is possible.

I assume you have gone through the following approach:

Split $\theta$ into $\alpha,\beta$ (such that $\alpha$ is above the X-axis, and $\beta$ is below the X-axis).

$\theta = \alpha + \beta$

Also,

$\tan\alpha = \frac{2at}{at^2} = \frac{2}{t}$

$\tan\beta = \frac{\frac{-2a}{t}}{\frac{a}{t^2}} = -2t$

I think you went wrong here. Maybe you assumed $\tan\beta$ to be positive (by taking the modulus)

Anish Puthuraya - 7 years, 2 months ago

@Anish Puthuraya and @Karthik Kannan , thank you both for your helpful replies. I think I see the error in my approach. As Karthik said, my formula does give the acute angle. The minimum acute angle from my formula comes out to be $\arctan(4/3)$ , this corresponds to maximum obtuse angle $\pi-\arctan(4/3)=126.87 ^{\circ}$ . This agrees with the given answer. Thank you once again.

Anish, it would be better to use ^{\circ} for the degree symbol. :)

Pranav Arora - 7 years, 2 months ago

The picture in the problem shows an angle not involving the parabola's vertex, which suggests (incorrectly) that as with a circle the angle made in this way is the same whichever point you choose.

Mark C - 5 years, 1 month ago

Bill Bell
Aug 18, 2015

I believe it's unnecessary to parameterise the parabola; it's sufficient to use $y={ x }^{ 2 }$ to make life simpler. However, I'm past the time in my life when I want to do any significant amount of calculus. And I didn't know the geometry presented elsewhere here.

from pylab import *
from math import pi, atan
from numpy import degrees
from scipy.optimize import minimize_scalar

def Theta(x):
    val=pi-(-atan(x**2/x)+atan((-1./(4*x))**2/(-1./(4*x))))
    return degrees(val)

x=[-.01*i for i in range(125,0,-1)]
theta=[Theta(t) for t in x]

plot(x, theta)

ylabel('theta (degrees)')
xlabel('x')
title('Two Points on a Parabola')
grid(True)
show()

f=lambda x: abs(130-Theta(x))
result=minimize_scalar(f,bounds=[-1.4,-0.1],method='bounded')
print result
print Theta(result.x)

Script result:

 status: 0
    nfev: 11
 success: True
     fun: 3.1301023541570601
       x: -0.49999986043913963
 message: 'Solution found.'
126.869897646

Fun question!

Two Points On A Parabola

The answer is 216.87.

2 solutions

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