# Find an equation of the tangent line to the curve 2(x^(2)+ y^(2))^(2)= 25(x^(2)-y^(2)) at the point(3,1).

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You need to remember that the slope of the curve is `(dy)/(dx), ` hence, you need to differentiate the functionwith respect to x such that:

`4(x^2 + y^2)(2x) + 4(x^2 + y^2)(2y)(dy)/(dx) = 50x - 50y(dy)/(dx)`

`-8x(x^2 + y^2) + 50x = (50y+ 8y(x^2 + y^2)) (dy)/(dx)`

You need to find (dy)/(dx) such that:

`(dy)/(dx) = (-8x(x^2 + y^2) + 50x )/(50y + 8y(x^2 + y^2))`

You need to substitute 3 for x and 1 for y in equation

`(dy)/(dx) = (-8x(x^2 + y^2) + 50x )/(50y + 8y(x^2 + y^2)) ` such that:

`(dy)/(dx)|_(3,1) = (-8*3(3^2 + 1^2) + 50*3 )/(50 + 8(3^2 + 1^2))`

`(dy)/(dx)|_(3,1) = (-240 + 150)/(50+80)`

`(dy)/(dx)|_(3,1) = -90/130 =gt (dy)/(dx)|_(3,1) = -9/13`

**Hence, evaluating the tangent line to the given curve, at the point (3,1), yields `(dy)/(dx)|_(3,1) = -9/13` .**