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### Tangent to Parabola

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Date: 10/21/96 at 22:37:4
From: Dave Spoelhof
Subject: Tangent to Parabola

Dr. Math,

Could you help me me figure out the slopes of two lines tangent
to the parabola y = x^2 which pass through the point (2,1)?
The line which passes through (2,1) and (0,0) with a slope of 1/2
seemed like a good bet, but I can't figure out a second tangent.
Thanks - this is for a 10th grade math class.
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Date: 10/22/96 at 8:12:27
From: Doctor Jerry
Subject: Re: Tangent to Parabola

Dear Dave,

I'm going to assume that solving quadratic equations is within bounds.
(There are methods based on calculus, but since you said 10th grade,
I'll assume that these are out of bounds for a while.)

All non-vertical lines through (2,1) have the form y - 1 = m(x - 2).
We're looking for values of the slope m for which the line will be
tangent to the parabola.  This means that the line will intersect the
parabola exactly once.  Thus, when we solve the system

y - 1 = m(x - 2)
y = x^2

we want just one solution.

Substituting from the second equation into the first gives

x^2 - 1 = m(x - 2).

Rearranging this expression gives

x^2 - mx + (2m - 1) = 0.

This quadratic will have exactly one real root if its discriminant
b^2 -4ac is zero.  Thus,

m^2 - 4*1*(2m - 1) = 0.

The roots of this quadratic are 4 + 2*sqrt(3) and 4 - 2*sqrt(3).

These are the slopes of the lines through (2,1) that can be tangent to
the given parabola.

The line you mentioned is not tangent to the parabola.  If you have a
graphing calculator, try graphing the parabola and line together, with
the range on x restricted to -2 to 2.  You'll notice that the line
crosses the parabola and is not tangent to it.

I hope this helps.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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```
Date: 10/22/96 at 21:47:21
From: Dave Spoelhof
Subject: Re: Tangent to Parabola

Thank you for the very complete and timely response!
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Associated Topics:
High School Basic Algebra
High School Conic Sections/Circles
High School Geometry

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