How do you find the point (in the first quadrant) on the lemniscate ##2(x^2+y^2)2=25(x^2−y^2)## where the tangent is horizontal?

The point on the lemniscate that has a horizontal tangent, in the first quadrant, is ##((5sqrt(3))/4, 5/4)##. Refer to the explanation

The lemniscate equation is ##2(x^2 + y^2)^2 = 25(x^2 - y^2)## Here are a few things you can extract from the question. I'll refer to this information when necessary (you'll know when):

• ##color(orange)["a horizontal tangent line means that the slope is zero"]##. Mathematically, ##y'= (dy)/dx = 0##.
• ##color(red)["first quadrant answers mean that all x and y values are positive"]##
• ##color(blue)["you'll need to implicitly differentiate"]##. Let ##y' = (dy)/dx##.

Step 1) The requires the and the . ##2(x^2 + y^2)^2 = 25(x^2 - y^2)## ##2(2)(x^2 +y^2)^1(2x + color(blue)(2yy')) = 25(2x-color(blue)(2yy'))## ##4(x^2 +y^2)(2x + 2yy') = 25(2x-2yy')## ##4(x^2+y^2)(2x + color(orange)0) = 25(2x - color(orange)0)## ##4(x^2+y^2)(2x) = 25(2x)## ##4(x^2 + y^2) = 25## ##x^2 + y^2 = 25/4##

Step 2) You can solve for either ##x## or ##y## to find the values of the coordinate. I'll solve for ##x##. ##x^2 + y^2 = 25/4## ##x^2 = 25/4 - y^2## ##x = +-sqrt(25/4 - y^2)##. Since the point lies in the first quadrant, ##color(red)[x = sqrt(25/4-y^2)]##

Step 3) Back to the original equation to solve for y: ##2(x^2 + y^2)^2 = 25(x^2 - y^2)## ##2([sqrt(25/4-y^2)]^2 + y^2)^2 = 25([sqrt(25/4-y^2)]^2 - y^2)## ##2(25/4-y^2 + y^2)^2 = 25(25/4-y^2 - y^2)## ##2(25/4)^2 = 25(25/4 - 2y^2)## ##(2*25*25)/(4*4*25) = 25/4 - 2y^2## ##25/8 = 25/4 - 2y^2## ##2y^2 = 25/4 - 25/8## ##16y^2 = 50 - 25## ##16y^2 = 25## ##y^2 = 25/16## ##y = +- sqrt(25/16)## ##color(red)[y = sqrt(25/16)]## ##y = 5/4##

Step 4) Back to the statement ##x = sqrt(25/4-y^2)##, ##x = sqrt(25/4-(5/4)^2)## ##x = sqrt(25/4-25/16)## ##x = sqrt((100-75)/16)## ##x = +-sqrt(75/16)## ##color(red)[x = (5sqrt(3))/4]##

The point on the lemniscate that has a horizontal tangent, in the first quadrant, is ##((5sqrt(3))/4, 5/4)##.