What is the tangent line to the point on the curve $2x^2 - y^2 = 1$ at $x=5$?

Question

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!

Answer 1

You are correct.

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