# Thread: the point tangent to line

1. ## the point tangent to line

At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis

2. ## Re: the point tangent to line

Originally Posted by rcs
At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis

What is the gradient of the x axis? Hint: It's horizontal...

3. ## Re: the point tangent to line

Sir that term i coudnt answer ... im sorry i couldnt get the hint

4. ## Re: the point tangent to line

Originally Posted by rcs
At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis
$\displaystyle y+xy'=2(1-x-y)(-1-y')$ now find a point $\displaystyle (x,y)$ so that $\displaystyle y'=0$.

5. ## Re: the point tangent to line

If you do not know what "gradient' (American English: "slope") of a line is, you should not be attempting a problem like this. Finding the derivative is far more advanced that the 'slope' or 'gradient' of a line.

6. ## Re: the point tangent to line

Originally Posted by rcs
At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis

You've posted a lot of algebra questions on this forum, and this question might be too advanced for you. How much calculus do you know?

7. ## Re: the point tangent to line

i guess basic for u but im trying to know all these things sir

8. ## Re: the point tangent to line

Originally Posted by rcs
i guess basic for u but im trying to know all these things sir
What the hell does that mean?
That is not even a well formed English sentence
If you cannot post an proper statement or question then post nothing !

9. ## Re: the point tangent to line

Originally Posted by rcs
i guess basic for u but im trying to know all these things sir
I don't think you're quite ready for derivatives and calculus, especially since you've been posting elementary algebra questions. You need to be proficient in algebra to do well in calculus.

10. ## Re: the point tangent to line

The solution would involve differentiating both sides with respect to x:

$\displaystyle y + x \frac{dy}{dx} = 2(1-x-y)(-1 - \frac{dy}{dx})$ (using chain rule)

Then set dy/dx = 0.

11. ## Re: the point tangent to line

thank you Plato. You have a very positive statement . Keep it up!