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### Right Triangle Inscribed in a Parabola

```
Date: 09/20/1999 at 21:06:33
From: Mary Kay Mohs
Subject: Geometry with trig

A right triangle RST with hypotenuse ST is inscribed in the parabola
y = x^2 so that R coincides with the vertex of the parabola. If ST
intersects the axis of the parabola at Q, then show that Q is
independent of the choice of right triangle.

```

```
Date: 09/21/1999 at 19:27:39
From: Doctor Lilla
Subject: Re: Geometry with trig

Dear Mary,

Thanks for writing to Dr. Math. This is a really interesting problem.
It is always challenging to work with parameters, especially when we
need to prove a statement. I assume you have a good picture and
understanding of the problem.

Let's start with the parabola. The equation of the parabola is y =
x^2, meaning that each point of the curve can be written in P(x,x^2)
form.

Now, what else do we know? Well, we know that SR is perpendicular to
RT, since they are two sides of the right triangle.

Let's pick S arbitrarily from the parabola, so let S be (X0,X0^2).
X0 is the only parameter we will use. Since R is at the origin,
R(0,0).

Our steps from here should be the following:

1. Express the coordinates of T with X0.

2. Find the equation for line ST (using X0 as parameter).

3. Find the point of line ST that is on the y axis.

4. The coordinates of this point should not contain X0 any more, but
they should be constants. Actually, since this point is on the
y-axis, its first coordinate is 0, we just need to find the
ordinate of this point.

1. The slope of SR is

X0^2
----  =  X0.
X0

The slope of RT, since RT is perpendicular to SR is

1
- --
X0

Since R(0,0) is on this line, using the point-slope form,

1
y  =  - -- * x
X0

x
=  - --
X0

Because T is on the parabola and on RT, we have to solve the system of
equations:

x
y  =  x^2     and     y  =  - --
X0

x
x^2  =  - --
X0

gives us 2 solutions:

(0,0)           that is R, and

1    1
(- --, ---- )   which is T.
X0  X0^2

2. We want to find the equation of ST, where S(X0,X0^2), T is above.

The slope of ST:

1
X0^2 - ----
X0^2          1
m  =  -----------  =  X0 - --     (can you see why?)
1             X0
X0 + --
X0

I want to leave something to you, so I'll let you find the equation of
line ST, and follow the steps above from here. Once you have that
equation, you have to find the point of ST that is on the y-axis
(having 0 as the x-coordinate).

If this value is a number, not a parameter, you have proved that no
matter how we choose point S (so no matter what X0 is), ST will always
cross the y-axis at that certain point.

Please write back if you have problems or more questions on this.

- Doctor Lilla, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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