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Consider the curve described by $2x^2 - y^2 = 1$. What is the equation of the tangent line to the point on the curve in the fourth quadrant with x-coordinate $x = 5$?

My solution:

Derivative by $x$: $4x-2yy'=0$, and the point in the 4th quadrant is $(5, -7)$, thus the slope is $y'=20/-14=-10/7$, the tangent line is $y+7=-\frac{10}{7}(x-5)$.(does not need to simplify)

Is it the correct answer (the test result said it was WRONG)?

Thanks!

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    $\begingroup$ Your answer is correct. ;) $\endgroup$
    – NasuSama
    Commented Jan 4, 2014 at 21:28

1 Answer 1

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You are correct.

(Posting so this doesn't end up on the unanswered questions queue.)

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