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anonymous
 5 years ago
Find the equation of the tangent line to the curve y2+4xy+x2 = 13
at the point (2; 1).
anonymous
 5 years ago
Find the equation of the tangent line to the curve y2+4xy+x2 = 13 at the point (2; 1).

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bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0find the first derivative: 2y*y' + 4(x*y' + y) + 2x = 0 2y*y' + 4x*y' + 4y + 2x = 0

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0all you have to do now is find y' as it is also the slope of the tangent line. Plug in 2 for x and 1 for y: 2*y' + 4*2*y'+4*1+2*2 = 0 2y'+8y' = 44 10y' = 8 y' = M = 8/10

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Now use the formula: y  y1 = M(xx1) y1 = (4/5)(x2)

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0M = 8/10 = 4/5 thats how i got 4/5 in the previous part..

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Simplify: y1 = (4/5)(x2) y = 4x/5 + 8/5 +1 y = 4x/5 + 13/5

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah that helped a lot! it's been a few years since i did this stuff so i totally forgot about using y' as an intermediate variable, makes sense now tho, thanks and I will fan you :)
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