# Tangent from a point to a curve?

• Jun 23rd 2012, 05:45 AM
NitroNbg
Tangent from a point to a curve?
I have an important exam on Monday and I stumbled upon this problem.
Problem: Calculate the angle (with tan(x) function) between the tangent lines drawn from (-4,1) point to a graph y^2=2*x.

I know that derivative of function represents its tangent line equation, however I cannot seem to implement the given point into the equation since it's out of the domain. Is the correct derivative:
Since the given equation represents a non "1-1" function, divide the graph into two separate graphs (y=sqrt(2*x) and y=-sqrt(2*x)) then dy=1/sqrt(2*x) and dy=-1/sqrt(2*x). I know I did something wrong but not sure what.

I hope somebody can help me. Thanks in advance
• Jun 23rd 2012, 06:48 AM
mfb
Re: Tangent from a point to a curve?
"dy=1/sqrt(2*x)" is wrong. I think you mean dx/dy=1/sqrt(2*x)

You have to find the points where the tangents touch the function yourself. Name these points, find equations which relate the slope of the tangent (at this point) to the condition that the tangent has to go through your point.

It should look similar to this:
Attachment 24142
• Jun 23rd 2012, 07:53 AM
richard1234
Re: Tangent from a point to a curve?
$\displaystyle y^2 = 2x$. Differentiating both sides with respect to x, $\displaystyle 2y \frac{dy}{dx} = 2 \Rightarrow \frac{dy}{dx} = \frac{1}{y}$.

This means that for a given $\displaystyle (x,y)$ on the function, the slope of the tangent line at that point is $\displaystyle \frac{1}{y}$. Since the tangent line also goes through (-4,1) we can write out an equation for the slope:

$\displaystyle \frac{1}{y} = \frac{y-1}{x-(-4)} = \frac{y-1}{x+4} \Rightarrow x+4 = y(y-1)$

We know that $\displaystyle x = \frac{y^2}{2}$, so

$\displaystyle \frac{y^2}{2} + 4 = y(y-1)$. Solve the quadratic.
• Jun 23rd 2012, 08:54 AM
Imo
Re: Tangent from a point to a curve?
First take a derivate of your function, it will be y'=1/sqrt(2*x)
Now write a tangent eg. over two points: T1(-4,1), T2(y2,x2) - point where tangent line touches our curve:
y2=√(2*x2 ), y-1= √2*x2 -1/(x2+4)(x+4)

Now, we know that √2*x2 -1/(x2+4) has to be equal 1/sqrt(2*x2) , and from that we can find x2=8.
y2=√2*×2_ =4.
Now u have two points to define ur tangent.
Same way find the other tangent and angle between them.
Hope this is correct.
• Jun 23rd 2012, 09:44 AM
NitroNbg
Re: Tangent from a point to a curve?
Your suggestions were correct. After 10+mins I finally got to the correct answer. It took me that long cuz I was going the long way. While on that topic - I know two possible solutions to calculate the required tan(x).
First one is to use tan(theta)=slope of the corresponding function and then use the identity tan(x+y)=(tan(x)+tan(y))/(1-tan(x)*tan(y)).

The other one is to find coordinates of the dots that are part of right angle triangle (dots being (-4,1), circle and one tangent line intersection (2,-2) and radius perpendicular to that point and other tangent line intersection (32/7,22/7)).

My question is - is there a shorter way?

After all the calculations, I got the correct answer which is arctan(6/7) if anyone is wondering.
Thanks Imo and richard1234