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Mnenie [13.5K]
4 years ago
7

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Mathematics
1 answer:
Nimfa-mama [501]4 years ago
6 0
\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}
\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}

Now, isolate the variables, so you can integrate.
(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}
\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}
2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C


4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C
y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C


y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}
y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}

At x = 1, y = 0.
0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}
-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}

4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0
\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}


4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C
0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C
C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}
C = -\frac{\pi}{2}

\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}
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Answer:  \bold{1.\quad \text{Vertex}:(4,-1)\qquad \text{Focus}:\bigg(4,-\dfrac{9}{8}\bigg)\qquad \text{Directrix}:y=-\dfrac{7}{8}}

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<u>Step-by-step explanation:</u>

The vertex form of a parabola is y = a(x - h)² + k  or  x = a(y - k)² + h

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    \bullet\quad a=\dfrac{1}{4p}

1) y = -2(x - 4)² - 1     →      a = -2   (h, k) = (4, -1)

a=\dfrac{1}{4p}\quad \rightarrow \quad -2=\dfrac{1}{4p}\quad \rightarrow \quad p=-\dfrac{1}{8}\\\\\text{Focus = Vertex + p}\\\\.\qquad = \dfrac{-8}{8}+\dfrac{-1}{8}\\\\.\qquad =-\dfrac{9}{8}\qquad \rightarrow \qquad \text{Focus}=\bigg(4,-\dfrac{9}{8}\bigg)\\\\\\\text{Directrix: y=Vertex - p}\\\\.\qquad \qquad y=\dfrac{-8}{8}-\dfrac{-1}{8}\\\\.\qquad \qquad y=-\dfrac{7}{8}

*******************************************************************************************

2) x = (y - 1)² + 2     →      a = 1   (h, k) = (2, 1)

a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=-\dfrac{1}{4}\\\\\text{Focus = Vertex + p}\\\\.\qquad = \dfrac{8}{4}+\dfrac{1}{4}\\\\.\qquad =\dfrac{9}{4}\qquad \rightarrow \qquad \text{Focus}=\bigg(\dfrac{9}{4},1\bigg)\\\\\\\text{Directrix: x=Vertex - p}\\\\.\qquad \qquad x=\dfrac{8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=\dfrac{7}{4}

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