Asked by Skye
Find the equation of the tangent line
5x^2+y^2=14 at (1,3).
P.S: It's calculus.
So far I know this much
10x+2y(dy/dx)=o
-10x -10x
___________________
2y(dy/dx)= -10x
(dy/dx)= -10x/2y= -5x/y
m= -5(1)/3= -5/3
y=mx+b
3= -5/3(1)+b then what?
5x^2+y^2=14 at (1,3).
P.S: It's calculus.
So far I know this much
10x+2y(dy/dx)=o
-10x -10x
___________________
2y(dy/dx)= -10x
(dy/dx)= -10x/2y= -5x/y
m= -5(1)/3= -5/3
y=mx+b
3= -5/3(1)+b then what?
Answers
Answered by
Reiny
ahh, so close!
solve for b in your last equation
I get b = 14/3
put m and b into your y = mx + b and you are done
solve for b in your last equation
I get b = 14/3
put m and b into your y = mx + b and you are done
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