SOLUTION: Find an equation of the tangent line to the graph at the given point. 7x^2-6(sq3)xy+13y^2-16=0. (sq3,1)
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Question 625669: Find an equation of the tangent line to the graph at the given point. 7x^2-6(sq3)xy+13y^2-16=0. (sq3,1)
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
7x^2-6(sq3)xy+13y^2-16 = 0
dy/dx = -(14x - 6sqrt(3)y)/(26y - 6sqrt(3)x)
------------------
f'(sqrt(3),1) = -(14sqrt(3) - 6sqrt(3))/(26 - 18)
= - sqrt(3) slope of tangent line
-------------
y = mx + b
1 = -3 + b
b = 4
--> y = -sqrt(3)*x + 4
==================
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Differentiate the equation implicitly:
\ =\ \frac{d0}{dx})
}{dx}\ -\ \frac{d\left(6\sqrt{3}xy\right)}{dx}\ +\ \frac{d\left(13y^2\right)}{dx}\ -\ \frac{d(16)}{dx}\ =\ 0)
Use the Chain Rule and the Product Rule:
\ +\ 13y\frac{dy}{dx}\ -\ 0\ =\ 0)
Now take out a common factor of 2 and then solve for 
Do not succumb to the temptation to rationalize your denominator at this point. We have an expression for the tangent to your ellipse at any point, all that is needed now is to evaluate this expression at the given point. Substitute the coordinates of the given point for 
and 
in the expression for and do the arithmetic.
John
My calculator said it, I believe it, that settles it
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