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 (Crying)

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- Mar 29th 2008, 12:28 PMlooi76[SOLVED] Differentiation Help!
**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 (Crying) - Mar 29th 2008, 12:44 PMwingless
**Hint:**$\displaystyle y=2x+c$ tells us the tangent lines slope is $\displaystyle m=2$. So y' is 2 at the tangent point. - Mar 29th 2008, 12:46 PMTrevorP
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. - Mar 29th 2008, 01:50 PMwingless
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.