# 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

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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?
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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)##.

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