Lines tangent to parabola at point. 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!
 A: 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.
A: 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 $y-4=4(x-2)$  
A: 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 $y=2^2$.
A: 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.
