Calculus w/ Analytical Geometry question!?

I have already learned methods of differentiations and integrations. However, it seems that i still do not have enough knowledge to be able to do this problem:

Q:

Find the coordinates of all points (x,y) on the graph of the equation x^2-xy+y^2 = 12 where the tangent is parallel to the y-axis.

My approach so far is to use implicit differentiation and plug in y=0, but i do not know what to do after this step.

Thank you if you can help me!

Update:

BY the way, the answer given is : (-4,-2) & (4,2)

2 Answers

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  • Hemant
    Lv 7
    9 years ago
    Best Answer

    ... x² - xy + y² = 12 ..... (1)

    ∴ diff... w.r.t. x by Chain Rule,

    ... 2x - ( x.y' + y.1 ) + 2y.y' = 0

    ∴ ( 2y - x ) y' = y - 2x

    ∴ y' = ( y - 2x ) / ( 2y - x )

    ∴ slope of normal = -1/ y' = - ( 2y - x ) / ( y - 2x ) ....... (2)

    Since tgt is parallel to Y-axis ∴ normal is parallel to X-axis.

    ∴ slope of normal = 0 ∴ from (2), ... 2y - x = 0 ∴ x = 2y

    ∴ from (1),

    ... (2y)² - (2y)y + y² = 12 ∴ 3y² = 12 ∴ y² = 4 ∴ y = ± 2

    ∴ x = 2y = 2 ( ± 2 ) = ± 4

    ∴ x = ± 4, y = ± 2

    ∴ the required points are ( x, y ) ≡ ( 4, 2 ), ( - 4, -2 ) .............. Ans.

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    Happy To Help !

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  • 9 years ago

    implicit differentiation ok

    plug in y = 0 wrong

    dy/dx and look for values of x that make the ans infinity

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