The angle between the curves 2x^2 + y^2 = 20 | vedclass

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(B) Given equations are $2x^2 + y^2 = 20$ $(1)$ and $4y^2 - x^2 = 8$ $(2)$.From $(2)$, $x^2 = 4y^2 - 8$. Substituting this into $(1)$:$2(4y^2 - 8) + y

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$\frac{\pi}{2}$

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(B) Given equations are $2x^2 + y^2 = 20$ $(1)$ and $4y^2 - x^2 = 8$ $(2)$.From $(2)$, $x^2 = 4y^2 - 8$. Substituting this into $(1)$:$2(4y^2 - 8) + y^2 = 20$$8y^2 - 16 + y^2 = 20$$9y^2 = 36 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.Since the intersection is in the $4^{th}$ quadrant, $y = -2$.Substituting $y = -2$ into $x^2 = 4y^2 - 8$:$x^2 = 4(4) - 8 = 8 \Rightarrow x = \pm 2\sqrt{2}$.In the $4^{th}$ quadrant, $x > 0$, so $x = 2\sqrt{2}$. The point is $(2\sqrt{2}, -2)$.Differentiating $(2)$: $8y \frac{dy}{dx} - 2x = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{4y}$.At $(2\sqrt{2}, -2)$, $m_1 = \frac{2\sqrt{2}}{4(-2)} = -\frac{\sqrt{2}}{4} = -\frac{1}{2\sqrt{2}}$.Differentiating $(1)$: $4x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{2x}{y}$.At $(2\sqrt{2}, -2)$, $m_2 = -\frac{2(2\sqrt{2})}{-2} = 2\sqrt{2}$.Since $m_1 	imes m_2 = (-\frac{1}{2\sqrt{2}}) 	imes (2\sqrt{2}) = -1$, the curves are orthogonal.Therefore, the angle between the curves is $\frac{\pi}{2}$.

The angle between the curves $2x^2 + y^2 = 20$ and $4y^2 - x^2 = 8$ at a point where they intersect in the $4^{th}$ quadrant is

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