Question:Find the value of the constant c for which the line $\displaystyle y = 2x+c$ is a tangent to the curve $\displaystyle y^2 = 4x$
Attempt:
NO IDEA
Well think about it. You are told that you are to find a line that is TANGENT to another line.
This means when the derivative of the second equation equals the slope of the first one:
$\displaystyle y^2 = 4x \rightarrow y = 2\sqrt{x}$
and you want when it is equal to 2:
$\displaystyle \frac{dy}{dx} = 2$
AND also you know that they both are equal at those x values. So you should be able to get two equations and solve them for C.
Not true.
Remember: $\displaystyle \sqrt{x^2}$ is NOT $\displaystyle x$, its $\displaystyle |x|$!
So,
$\displaystyle y^2 = 4x$
$\displaystyle \sqrt{y^2} =\sqrt{4x}$
$\displaystyle |y| = \sqrt{4x} = 2\sqrt{x}$
This gives us two functions of y:
$\displaystyle y= 2\sqrt{x}$ and $\displaystyle y= -2\sqrt{4x}$.
Now differentiate both and set both derivatives y' = 2. Only one of them will satisfy it and you'll get one value for x. This is the x coordinate of the tangent point. Find its y coordinate too and you'll get the tangent point. Plug this point in the line equation to get c.
A better approach is here:
If $\displaystyle \frac{dy}{dx} = 2$, then $\displaystyle \frac{dx}{dy} = \frac{1}{2}$
$\displaystyle y^2 = 4x$
$\displaystyle x' = \frac{y}{2}$
Now set $\displaystyle x'=2$ to get the tangent points y coordinate. This is much easier than working with two functions.