Prove: The chord joining the points of contact of the tangent lines to a parabola from any point on its directrix passes through its focus.

Question

Hello, struggling with this one and appreciate help!

Prove: The chord joining the points of contact of the tangent lines to a parabola from any point on its directrix passes through its focus.

For this chapter/topic, I somehow need to show the points of contact between the tangent lines that intersect at some point (x₁,-p) on the directrix, and then show that the equation of the line between these two points passes through the focus (0,p).

If I use the parabola 4y=x², and take the derivative (x/2), I get the tangent line equation (y-y₀)=(x/2)(x-x₀). I can't figure out how to use point of intersection (x₁,-p) on the directrix and the equation of the tangent line to find the two points of tangency on the parabola.

Thank you for your help!!

Answer 1

Hi Andrew, this is so much easier to explain "live". Please see video.

Sorry it is so long. It is just a lot of explaining.

Answer 2

I will add to Doug C.'s information.

I can't prove this, but I believe it to be true.

If the tangent a x₀ intersects the directrix at point p, the other tangent from point p will be perpendicular to the first tangent.

.I will work on this problem some more and get back to you if I have any more help.

Answer 3

Here is a start forthis, where f(x) = 1/4 x².

If (x₀, f(x₀)) if a point on the parabola that is the POT for a tangent line passing through (x₁, -p), then the slope of the line joining those two points (by the slope formula) must be the same as f'(x₀).

That is: [f(x₀) - (-p)]/ (x₀-x₁) = 1/2 x₀

_{Here is a Desmos graph illustrating this with f(x) = 1/12 x}² _{(so p = 3)}

desmos.com/calculator/vvrawhtvwe

Convince yourself that the conjecture is true for a specific example (which you are in the process of doing), then move to a general proof -- I bet this will get messy.

Answer 4

For y = x^2/4, the focus is at (0, 1) and the directrix is at y = -1.

The tangent at an arbitrary point on the parabola (x_o,x_o^2/4) has slope x_o/2.

Thus, the equation for the tangent is x_o^2/4 = x_o/2.x_o + b, so b = - x_o^2/4.

Thus, as the equation of the line is y = x_o/2 x - x_o^2/4, when y = -1,

-1 = x_o/2 x - x_o^2/4, or -1 + x_o^2/4 = x_o/2 x, or x = -2/x_o + x₀/2

We may rewrite this as x₀/2 -2/x_o - x = 0, or x₀^{^2} - 2 x_o x - 4 = 0

x_o = x +- √(x^2 + 4)

Then, as y = x^2/4, we have yo = x^2/2 + 1 + 1/2x√(x^2 + 4) when x_o = x + √(x^2 + 4) and

yo = x^2/2 + 1 - 1/2x√(x^2 + 4) when x_o = x - √(x^2 + 4)

Then, the slope of the line between the 2 points will be (x^2/2 + 1 + 1/2x√(x^2 + 4) - ( x^2/2 + 1 - 1/2x√(x^2 + 4))/(x + √(x^2 + 4) - (x + √(x^2 + 4) )) = x√(x^2 + 4)/2√(x^2 + 4) = x/2

Then, as we have the point (x + √(x^2 + 4),x^2/2 + 1 + 1/2x√(x^2 + 4)) and the slope is x/2

use y = mx + b

x^2/2 + 1 + 1/2x√(x^2 + 4) = x/2((x + √(x^2 + 4)) + b

x^2/2 + 1 + 1/2x√(x^2 + 4) = x^2/2 + 1/2x √(x^2 + 4) + b

1 = b

Then, when x = 0, y = mx + b = m(0) + 1 = 1

(0,1) is the focus. Thus, the line between the two tangents to a point on the directrix goes through the focus.