Hearts And Conics.

Calculus Level pending

Find the conic a x 2 + + b x y + c x 2 + d x + e y = 1 ax^2 + + b xy+ cx^2 + dx + ey = -1 , using the five points below, that intersects the curve x 2 + ( 5 y 4 x ) 2 = 1 x^2 + \left(\frac{5y}{4} - \sqrt{|x|}\right)^2 = 1 .

Then, to six decimal places, find the area of the region bounded by the curve f ( x ) = 4 5 ( x 1 x 2 ) f(x) = \frac{4}{5}\big(\sqrt{x} - \sqrt{1 - x^2}\big) , the positive x x -axis, the lines y = 4 5 y = \frac{4}{5} and y = 4 5 , y = -\frac{4}{5}, and the conic.


The answer is 0.091559.

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1 solution

Rocco Dalto
Dec 14, 2019

To show the conic is an ellipse.

Using the point ( 1 , 4 5 ) (1,\dfrac{4}{5}) we obtain (1): a + 4 5 b + 16 25 c + d + 4 5 e = 1 a + \dfrac{4}{5}b + \dfrac{16}{25}c + d +\dfrac{4}{5}e = -1

Using the point ( 0 , 4 5 ) (0,\dfrac{4}{5}) we obtain (2): 16 25 c + 4 5 e = 1 \dfrac{16}{25}c + \dfrac{4}{5}e = -1

Using the point ( 0 , 4 5 ) (0,\dfrac{-4}{5}) we obtain (3): 16 25 c 4 5 e = 1 \dfrac{16}{25}c - \dfrac{4}{5}e = -1

Using the point ( 5 1 2 , 0 ) (\dfrac{\sqrt{5} - 1}{2},0) we obtain (4): ( 5 1 2 ) 2 a + ( 5 1 2 ) d = 1 (\dfrac{\sqrt{5} - 1}{2})^2a + (\dfrac{\sqrt{5} - 1}{2})d = -1

Using the point ( 5 1 2 , 0 ) (\dfrac{\sqrt{5} - 1}{2},0) we obtain (4): ( 5 1 2 ) 2 a ( 5 1 2 ) d = 1 (\dfrac{\sqrt{5} - 1}{2})^2a - (\dfrac{\sqrt{5} - 1}{2})d = -1

Using (4) and (5) we obtain: a = ( 2 5 1 ) 2 a = -(\dfrac{2}{\sqrt{5} - 1})^2 and d = 0 d = 0 .

Using (2) and (3) we obtain: c = 25 16 c = \dfrac{-25}{16} and e = 0 e = 0 .

Using (1) b = 5 4 ( 2 5 1 ) 2 \implies b = \dfrac{5}{4} (\dfrac{2}{\sqrt{5} - 1})^2 .

Let x 0 = 5 1 2 a = 1 x 0 2 , b = 5 4 ( 1 x 0 2 ) , c = 25 16 x_{0} = \dfrac{\sqrt{5} - 1}{2} \implies a = \dfrac{-1}{x_{0}^2}, b = \dfrac{5}{4}(\dfrac{1}{x_{0}^2}), c = \dfrac{-25}{16} , and d = e = 0 16 x 2 20 x y + 25 x 0 2 y 2 = 16 x 0 2 d = e = 0 \implies \boxed{16x^2 - 20xy + 25x_{0}^2y^2 = 16x_{0}^2} and b 2 4 a c < 0 b^2 - 4ac < 0 \implies we have an ellipse.

To find the area desired we need to solve for y = g ( x ) y = g(x) :

Solving for y y we obtain y = 2 5 x 0 2 ( ± 4 x 0 4 ( 4 x 0 2 1 ) x 2 + x ) y = \dfrac{2}{5x_{0}^2}(\pm\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2} + x) .

We want the ellipse below the line y = 2 5 x 0 2 x y = \dfrac{2}{5x_{0}^2}x which is g ( x ) = 2 5 x 0 2 ( 4 x 0 4 ( 4 x 0 2 1 ) x 2 + x ) g(x) = \dfrac{2}{5x_{0}^2}(-\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2} + x)

Using the given f ( x ) = 4 5 ( x 1 x 2 ) f(x) = \dfrac{4}{5}(\sqrt{x} - \sqrt{1 - x^2}) \implies

A 1 = x 0 1 g ( x ) f ( x ) d x A_{1} = \int_{x_{0}}^{1} g(x) - f(x) dx and A 2 = 0 x 0 f ( x ) g ( x ) d x A_{2} = \int_{0}^{x_{0}} f(x) - g(x) dx .

For A 1 A_{1} :

Let A = x 0 1 2 5 x 0 2 ( 4 x 0 4 ( 4 x 0 2 1 ) x 2 + x ) d x A^{*} = \int_{x_{0}}^{1} \dfrac{2}{5x_{0}^2}(-\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2} + x) dx

Let 4 x 0 2 1 x = 2 x 0 2 sin ( θ ) d x = 2 x 0 2 4 x 0 2 1 d θ \sqrt{4x_{0}^2 - 1} x = 2x_{0}^2 \sin(\theta) \implies dx = \dfrac{2x_{0}^2}{\sqrt{4x_{0}^2 - 1}} d\theta

A = 2 5 x 0 2 ( 2 x 0 4 4 x 0 2 1 ( arcsin ( 4 x 0 2 1 x 2 x 0 2 ) x 2 4 x 0 4 ( 4 x 0 2 1 ) x 2 ) + x 2 2 ) x 0 1 . 135739 \implies A^{*} = \dfrac{2}{5x_{0}^2}(-\dfrac{2x_{0}^4}{\sqrt{4x_{0}^2 - 1}} * (\arcsin(\dfrac{\sqrt{4x_{0}^2 - 1}x}{2x_{0}^2}) - \dfrac{x}{2}\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2}) + \dfrac{x^2}{2})|_{x_{0}}^{1} \approx \boxed{.135739} .

Let A = x 0 1 4 5 ( x 1 x 2 ) d x A^{**} = \int_{x_{0}}^{1} \dfrac{4}{5}(\sqrt{x} - \sqrt{1 - x^2}) dx

Let x = sin ( θ ) d x = cos ( θ ) d θ x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies

A = 4 5 ( π 4 1 2 arcsin ( x 0 ) 1 2 x 0 1 x 0 2 + 2 3 x 0 3 2 2 3 ) . 106728 A^{**} = \dfrac{4}{5}(\dfrac{\pi}{4} - \dfrac{1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - x_{0}^2} + \dfrac{2}{3} x_{0}^{\dfrac{3}{2}} - \dfrac{2}{3}) \approx \boxed{-.106728} .

A 1 = A + A . 029011 A_{1} = A^{*} + A^{**} \approx \boxed{.029011}

Similarly, for A 2 A_{2} we obtain:

A 2 = 4 5 ( 2 3 x 0 3 2 1 2 arcsin ( x 0 ) x 0 1 x 0 2 ) ) + 4 x 0 2 5 4 x 0 2 1 arcsin ( 4 x 0 2 1 2 x 0 ) . 062548 A_{2} = \dfrac{4}{5}(\dfrac{2}{3}x_{0}^{\dfrac{3}{2}} - \dfrac{1}{2}\arcsin(x_{0}) - x_{0}\sqrt{1 - x_{0}^2})) + \dfrac{4x_{0}^2}{5\sqrt{4x_{0}^2 - 1}}\arcsin(\dfrac{\sqrt{4x_{0}^2 - 1}}{2x_{0}}) \approx \boxed{.062548}

\implies The desired area is A = A 1 + A 2 = 0.091559 A = A_{1} + A_{2} = \boxed{0.091559} .

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