At point (1,2) of the curve x^2-xy + y^2 = 3
find the equation of the tanget line

Question

At point (1,2) of the curve x^2-xy + y^2 = 3
find the equation of the tanget line

aivantettet26 · Answer

tangent*

terenzreignz · Answer

You know tangent and derivative go together, right? Find
$$\Large \frac{dy}{dx}$$

terenzreignz · Answer

It's not for anything that tough, just for finding the slope of the line. Now come, find the derivative... implicitly (as if there's any other way)

aivantettet26 · Answer

2x+ y / x =y'?

terenzreignz · Answer

Whoa... I haven't even started yet... hang on XD

terenzreignz · Answer

$$\Large y^2-xy=3-x^2$$

aivantettet26 · Answer

hahaha! XD no prob!

terenzreignz · Answer

$$\Large 2y \color{red}{y'}-x\color{red}{y'}-y= -2x$$

terenzreignz · Answer

$$\Large \color{red}{y'}(2y - x )=y-2x $$

terenzreignz · Answer

Did I make a mistake?

aivantettet26 · Answer

but the answer is 2...n.n

anonymous · Answer

i see..is it only 2

terenzreignz · Answer

Wait what

aivantettet26 · Answer

y = 2

terenzreignz · Answer

Right, you made an error in your differentiation :P

aivantettet26 · Answer

xy is product rule right?

terenzreignz · Answer

This seems to hold:
$$\Large \color{red}{y'}=\frac{2y - x }{y-2x}$$

terenzreignz · Answer

Yup.
That's why differentiating it yields
$$\Large x\color{red}{y'}+ y$$

anonymous · Answer

ya u're ryte...

terenzreignz · Answer

So, the only thing left to do is substitute the values (1,2) to the derivative (which happens to be the slope of the tangent line at that point.
$$\Large \color{red}{y'}=\frac{2y - x }{y-2x}$$

aivantettet26 · Answer

find the equation of the tangent line

terenzreignz · Answer

Find the slope. Then we'll do the rest.
Just substitute x = 1 and y = 2 to $$\Large \color{red}{y'}=\frac{2y - x }{y-2x}$$

aivantettet26 · Answer

4/3

terenzreignz · Answer

$$\Large \color{}{y'}=\frac{2\color{red}x - \color{blue}y }{\color{red}x-2\color{blue}y}$$

terenzreignz · Answer

Sorry, typo.

terenzreignz · Answer

Redo it, substitute x = 1 and y = 2 in 
$$\Large \color{}{y'}=\frac{2\color{red}x - \color{blue}y }{\color{red}x-2\color{blue}y}$$

aivantettet26 · Answer

y' = 2(1) - 2 / 1-2(2)

terenzreignz · Answer

yes... go on :P

aivantettet26 · Answer

0?

terenzreignz · Answer

That's right :)
The slope is zero at that point.
It can only mean we have a horizontal tangent line :P

So... what is the equation of the HORIZONTAL LINE through the point (1,2) ?

aivantettet26 · Answer

y = 2!

terenzreignz · Answer

Correcto :P

anonymous · Answer

so if use y=mx+c method 2=0(1)+c
                                                c =2
                               so y=2
Thanks terenzreignz coz just now i read the question wrongly :(

terenzreignz · Answer

You read it wrongly, I answered wrongly, we're even @Asha_Lvnr ^_^

aivantettet26 · Answer

@Asha_Lvnr @terenzreignz  thank you for your efforts!

aivantettet26 · Answer