Last one!...find the points where vertical and horizontal tangents exisit. 4x^2 =y^2..

\[4x ^{2}+y ^{2}-8x+4y+4=0\]

what do you mean tangents exist?

in what context was this problem in?

that was exactly all it said..it was in chap of derivatives thou

then i'm guessing you have to find the derivative of the equation first... then, you make the numerator of the answer equal to zero and solve for that to find the horizontal tangent. then, do the same to the denominator to find the vertical tangent

i'm not 100 % positive though

ohhok..its fine..thnks 4 the attempt:))

the question is asking where dy/dx = 0, and where dx/dy = 0. for the first, take directive wrt x. then you get 8x+2yy'-8+4y'=0, since y' (or dy/dx) =0, you get x=1. when x=1, y=0, -4. for the second, take directive wrt y. then you get 8xx'+2y-8x'+4=0. since x' (or dx/dy) = 0, you get y=-2. when y=-2, you can solve the eq and get x=3, -1.

graphically, this is an ellipse. so points the question asks for are the four ends along the long and short axis. the equation can be also written as: (x-1)^2+(y+2)^2/4=1 the rest should be clear.

either way, you should be able to get to the correct answer.

oh wow..tht makes a lot of sense ! thank u so much!:)

sure.

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