find the lines that are tangent and normal to the curve at the given point:
x^2 + xy - y^2 = 1, (2,3)

Question

find the lines that are tangent and normal to the curve at the given point:
x^2 + xy - y^2 = 1, (2,3)

anonymous · Answer

looks like an implicit diff question 
you know how to do that?

anonymous · Answer

Kind of

anonymous · Answer

for the middle one you need the product rule
that is the only tricky part so remember

anonymous · Answer

$$x^2+xy-y^2=1\
2x+xy'+y-2yy'=0$$

anonymous · Answer

now you need $y'$ at $(2,3)$ 
you have a choice
you can solve for $y'$ using algebra, then plug in $(2,3)$ 
or you can plug in $x=2,y=3$ first  and solve for $y'$ that way

anonymous · Answer

y' should equal -3/2, right?

anonymous · Answer

i don't know, i didn't do it
you want me to check?

anonymous · Answer

I guess. I just algebra-d the relation out.

anonymous · Answer

$$4+2y'+3-6y'=0$$ would be my first step

anonymous · Answer

I didn't do that.

anonymous · Answer

i get $\frac{7}{4}$ we can try it the other way, really makes no difference

anonymous · Answer

Nope. I got it wrong. Again

anonymous · Answer

$$2x+xy'+y-2yy'=0$$
$$xy'-2yy'=-2x-y$$
$$y'(x-2y)=-2x-y$$
$$y'=\frac{-2x-y}{x-2y}$$ or $$y'=\frac{2x+y}{2y-x}$$

anonymous · Answer

then plug in the numbers, i also get $\frac{7}{4}$

anonymous · Answer

Ok, well I know 7/4. Now what do I do?

anonymous · Answer

use the point slope formula to find the equation of the line though $(2,3)$ with slope $\frac{7}{4}$

anonymous · Answer

that is the tangent line
the "normal" line is perpendicular, i.e. the slope is $-\frac47}{7}$

anonymous · Answer

oops $$-\frac{4}{7}$$

anonymous · Answer

so it's y-3 = 7/4(x-2)
y-3 = -4/7(x-2)