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I'm struggling to figure out what I'm exactly required to do. The problem states

"Compute which lines through the point $(1, 0)$ that are tangent to the parabola defined by $y = x^2$."

I believe it's a simple question however I've been going around this for quite a bit.

I'll appreciate any kind of help!

Thank you!

Matthew Conroy
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Joel Hernandez
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  • Draw the picture -- one of the tangent lines through $(1,0)$ should jump right out of the picture if your picture includes both axes. There is one other tangent line through $(1,0)$, but finding it takes a bit more work. – user128390 May 22 '14 at 21:17
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    Do it the other way: establish the general equation of a tangent to the parabola (through $(p, p^2)$, with slope $2p$) and express that it contains the point $(1, 0)$. –  May 22 '14 at 21:18
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    Write the equation of the tangent to the graph of the parabola $ y=x^2 $ at the generic point $(a, a^2) $. Which of these tangent lines go through the given point $(1, 0) $? – Américo Tavares May 22 '14 at 21:20
  • No matter how I try to express the general equation of the tangent to the parabola, I don't seem to see any intuition on this.. Could you attempt to explain me further ? I would truly appreciate it... – Joel Hernandez May 22 '14 at 23:49

4 Answers4

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Some comments were really helpful but I believe none gave me the intuition for this, which is what I really didn't understand , however I'm truly grateful for everyone that commented or attempted to help me.

Here's my approach after dreaming about it the entire night. (Not just the intuition)

We will first determine a general equation for all lines tangent to the function $f(x)$ at any given x , to do so let's create a generic point for our lines, let's make the point $(a,f(a))$ or equivalently $(a,a^2)$

We proceed to get the slope at any given value by getting the $d/dx$ of our function $f(x)$ wich results to be $f'(x) = 2x$

Next, let's evaluate the slope at our generalized point $(a,a^2)$

$$f'(a) = 2a$$

By doing this we can start to construct our line equation with $2a$ as the slope.

$$y=2ax+b$$

We're missing the b value, and to solve for it we will use the point that we are sure lie on our line equation, that is our point $(a,a^2)$ , so let's substitute it into our equation.

$$a^2=2a(a)+b$$

which is equal to

$$a^2=2a^2+b$$

Next, let's move the non b terms to the left , resulting $$-a^2=b$$

With that we can substitute our b to our first equation, resulting

$$y=2ax-a^2$$

This is the tricky part, we need to find which values of a would contain the point (1,0) (On this specific example) so we can substitute x = 1 and y = 0 on our equation, resulting

$$0=2a(1)-a^2$$

equivalently

$$0=2a-a^2$$

We proceed to factor out the a as

$$0=a(2-a)$$

We know that if our outcome it's zero then one of the terms being multiplied is going to be equal to zero, so we can say that either

$$a=0$$

or the term

$$2-a=0$$

Which we can easily solve for a resulting in

$$-a=-2$$

or $$a=2$$

so either a is equal to zero or a is equal to 2 , so we can substitute our "a's" on our first defined function for all tangent lines $y=2ax-a^2$, resulting in the line equations

$$y=2(2)x-(2^2)$$

Which is equivalent to $$y=4x-4$$

and $$y=2(0)x-(0^2)$$

Which is equivalent to $$y=0$$

So by doing these things we managed to get the equation of the lines tangent to $f(x)=x^2$ that go through the point (1,0)

I hope that this helps someone!

Regards.

Joel Hernandez
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2

The equation of a tangent line to the parabola at $(x_0,x_0^2)$ is given by $y-x_0^2=2x_0(x-x_0).$ If $(1,0)$ is a point of this line we have $-x_0^2=2x_0(1-x_0),$ or equivalently, $x_0^2=2x_0.$ So $x_0=0$ or $x_0= 2.$ That is, $(1,0)$ lies in the tangents to the parabola at points $(0,0)$ and $(2,4).$ Their equations are $y=0$ and $y-4=4(x-2)$

mfl
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  • ... or equivalently, $ x_0^2=2x_0 $. – Américo Tavares May 22 '14 at 21:35
  • I'm doing fine until the "or equivalently" , I've been trying all this time but can't seem to get the 0 or 1 result. Thanks for your answer anyway ! – Joel Hernandez May 22 '14 at 22:01
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    I have corrected the mistake found by Américo Tavares. To get the values $0$ or $2$ you have to solve the equation $x_0^2-2x_0=0,$ that is, $x_0\cdot (x_0-2)=0.$ – mfl May 22 '14 at 22:14
  • Hours later still on the same thing, I don't quite get the intuition on this.. Trying as hard as I can.. – Joel Hernandez May 22 '14 at 23:19
  • Hello Manuel, I believe that the equation that you mentioned on the answer at the end it's wrong as it doesn't touch the point 1,0 at any point, I graphed it, however the equation that did was y=4x-4 – Joel Hernandez May 23 '14 at 00:00
  • You are right. When I edited to correct the point in the equation I forgot to change the slope. The answer is $y-4=4(x-2).$ – mfl May 23 '14 at 08:19
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Certainly the $x$-axis is one such tangent line, since it passes through the point $(1,0)$ and is tangent to the parabola at its vertex. That much you get just from looking at the picture.

Classical geometry says that if you draw lines parallel to the axis of a parabola through the point of intersection of two tangent lines and through the two points of tangency, then the middle line (through the point of intersection) is exactly halfway between the other two lines parallel to the axis (through the points of tangency).

The $x$-coordinate of the point of intersection is $1$.

The $x$-coordinate of one of the points of tangency is $0$.

So $1$ is halfway between $0$ and what? Clearly $2$.

So the other tangent line is tangent to the parabola at $x=2$ and $y=2^2$.

Michael Hardy
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We can use basic "algebra." First draw a picture. That will pick up the fact that the $x$-axis is such a tangent line, and that there appears to be another one.

Now for the algebra. The tangent line "kisses" the parabola, so meets it at a "double-point."

The generic (non-vertical) line through $(1,0)$ has equation $y=m(x-1)$. We want the equation $m(x-1)=x^2$ to have a double root. The equation can be rewritten as $x^2-mx+m=0$. This has a double root if the discriminant $m^2-4m$ is equal to $0$, which happens at $m=0$ (which we knew) and $m=4$.

Remark: The above method goes back to Fermat, and a little later, Descartes.

André Nicolas
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