# Finding equation of tangent line

1. Dec 10, 2016

### Dank2

1. The problem statement, all variables and given/known data
The following point (x0,y0), is on the curve sqrtx +sqrty = 1

Show that line equation of the tangent line in the point. (x0,y0)
Is x/sqrtx0 + y/sqrty0 = 1

I've found the slope which is
-sqrty/sqrtx.
So slope of the point is -sqrty0/sqrtx0

2. Relevant equations

3. The attempt at a solution
y-y0=-sqrty0/sqrtx0(x-x0)

Should I get the solution from the equation above?

2. Dec 10, 2016

### Christofferk

Maybe it's easier if you model the problem as one of "one" variable. If you write it as a function y of x, then it is $y=x-2\sqrt{x}+1$, do you know what the tangent if this is at an arbitrary point? Try to look at the equation for the tangentplane of a $\textbf{R}^3$-function. The tangent-plane for such an equation at an arbitrary point $(x_0,y_0,z_0)$ is $f_{tangent}=\nabla f(x_0,y_0,z_0)\cdot(x-x_0,y-y_0,z-z_0)=0$

3. Dec 10, 2016

### Dank2

Haven't det with R3 yet. All we have learned is finding the slope and then using the formula y-y0 = m(x-x0)

4. Dec 10, 2016

### Christofferk

SOunds as if you know what you're doing, you just need to put the pieces of the puzzle together. :)

5. Dec 10, 2016

### Dank2

I think need a hint,

6. Dec 10, 2016

### Dank2

Is it like rw
is it like really easy to show from the equation I've written? I'm not sure which of the 4 variables should I substitute , I can't see how to get even 1 there

7. Dec 10, 2016

### Ray Vickson

You do not have 4 variables; you have only 2--namely x and y. These are the two objects that vary along the curve and along the tangent line. The other two objects ($x_0,y_0$) are just some constant input parameters. They do NOT vary along the curve or along the tangent line.

8. Dec 10, 2016

### Staff: Mentor

Thread moved. @Dank2, questions involving derivatives should NOT be posted in the Precalc section.

9. Dec 10, 2016

### Dank2

I meant I do not know which constant or variable should I subtitude in order to get the right form, from the main equation y= (1-sqrtx)^2 , y0=(1-sqrtx0)^2

Last edited by a moderator: Dec 10, 2016
10. Dec 10, 2016

### Staff: Mentor

IMO, and this is not a helpful hint.

11. Dec 10, 2016

### Staff: Mentor

The slope at the point $(x_0, y_0)$ is $\frac{-\sqrt{y_0}}{\sqrt{x_0}}$.
You know the point -- $(x_0, y_0)$ -- and you know the slope of the tangent line there -- $\frac{-\sqrt{y_0}}{\sqrt{x_0}}$. Finding the equation of a line through a given point and with a known slope should be straightforward.

12. Dec 11, 2016

### Dank2

Yes it is (y-y_0)=−√y0/√x0*(x-x_0). but i need to show it equals :
x/√x_0 + y/√y_0 = 1

Last edited: Dec 11, 2016
13. Dec 11, 2016

### Dank2

(y-y_0)=−√y0/√x0*(x-x_0) ==> y-y_0 = −√y_0*x/√x0 −√y_0*x_0/√x0
i think that somehow i need to use y=x−2√x+1 from the main equation, i just struggle on this point to do the algebra

14. Dec 11, 2016

### Ray Vickson

You have $\sqrt{y_0}$ in your first equation and $1/\sqrt{y_0}$ in your second equation. How, then, can you get from the first to the second?

15. Dec 11, 2016

### Dank2

1/√y_0 ? can't see it

Last edited: Dec 11, 2016
16. Dec 11, 2016

### Dank2

i need a hint

17. Dec 11, 2016

### Ray Vickson

You said you wanted to verify that $x/\sqrt{x_0} + y/\sqrt{y_0} = 1$.

18. Dec 11, 2016

### Dank2

that the tangent line equation at the point (x_0, y_0) is equal to it yes.

19. Dec 11, 2016

### Dank2

i need to proof it

20. Dec 11, 2016

### Dank2

i tried coming from the general equation of the tangent line by slope and point, and it didn't come out good.