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Calculus/Slope of a tangent.

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Expert: - 3/10/2011


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I have attached a question form calculas homework let me know if you can't see it). Could you please help me with it? I don't get it at all. Please explain as you go along. Thanks!

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Questioner:    Emily

I have attached a question form calculUs homework.  Could you please help me with it? I don't get it at all. Please explain as you go along. Thanks!  



I have moved your picture around so that the parabola points up instead of right.  So it is y = x^2 instead of x = y^2, and the point is (0,a) instead of (a,0).  All the logic is identical.
.......................................................
Normal from the point (0,a) to the parabola y = x^2?

At the point (x0,y0), which is NOT (0,0) we will have:

Slope of the tangent = dy/dx = 2x0

Slope of the normal line is -1/2x0  << negative reciprocal of slope of tangent.

But that is also the slope of the line between (0,a) and (x0,y0):
   y0 - a
m = ------
   x0 - 0

So:
y0 - a     -1
------ = -----
 x0      2x0


x0^2 - a      -1
---------- = -----
   x0        2x0

x0^2 - a      -1
---------- = -----
   1           2   << x0's cancel

2x0^2 - 2a = - 1   << cross-mult

2x0^2 + 1 =  2a    << move around terms

a = x0^2 + 1/2

But x0^2 > 0,  <<< because the point (x0,y0) is NOT (0,0), remember?

So:  a > 1/2

Now reversing it, if  a > 1/2, we will be able to :

solve a = x0^2 + 1/2  for x0

and find the three points we want.
....................................
Now, as to finding x0 such that the other two tangents (other than the one at the origin, this means) are perpendicular:

We solve a = x0^2 + 1/2  for x0 :

x0^2 = a - 1/2

x0 = +- sqrt(a - 1/2)

Write

x01 = + sqrt(a - 1/2)
x02 = - sqrt(a - 1/2)

Now the:

Slope of the normal line is -1/2x0  <<< did I say that before?

so the two slopes are:

 -1
-----
2x01

and


 -1
-----
2x02

and they are to be negative reciprocals:

 -1    2x02
----- = -----
2x01     1

So it must be true that:

4 x01 x02 = -1

Now if:
x01 = + sqrt(a - 1/2)
x02 = - sqrt(a - 1/2)

4(+ sqrt(a - 1/2))(- sqrt(a - 1/2)) = -1

Let's solve that for a:

-4 (a - 1/2) = -1

4 (a - 1/2) = 1

4a - 2 = 1

4a = 3

a = 3/4

I think that does it.

Cute problem.

Paul Klarreich

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All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

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