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A curve has equation $$x^2y^2 + xy^4 = 12$$ 9 (a) Prove that the curve does not intersect the coordinate axes - AQA - A-Level Maths Pure - Question 9 - 2019 - Paper 3

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A curve has equation $$x^2y^2 + xy^4 = 12$$ 9 (a) Prove that the curve does not intersect the coordinate axes. 9 (b) (i) Show that $$\frac{dy}{dx} = \frac{2xy + y... show full transcript

Worked Solution & Example Answer:A curve has equation $$x^2y^2 + xy^4 = 12$$ 9 (a) Prove that the curve does not intersect the coordinate axes - AQA - A-Level Maths Pure - Question 9 - 2019 - Paper 3

Step 1

Prove that the curve does not intersect the coordinate axes.

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Answer

To prove that the curve does not intersect the coordinate axes, we need to show that it does not satisfy the equations for the x-axis and y-axis.

For the x-axis: On the x-axis, we have (y = 0). Substituting (y = 0) into the equation: x2(0)2+x(0)4=120=12x^2(0)^2 + x(0)^4 = 12 \\ 0 = 12 This results in a contradiction, indicating that the curve does not intersect the x-axis.

For the y-axis: On the y-axis, we have (x = 0). Substituting (x = 0) into the equation: (0)2y2+0(y4)=120=12(0)^2y^2 + 0(y^4) = 12 \\ 0 = 12 Similarly, this also results in a contradiction. Therefore, the curve does not intersect the y-axis either.

In conclusion, since both intersections yield contradictions, the curve does not intersect the coordinate axes.

Step 2

Show that \( \frac{dy}{dx} = \frac{2xy + y^3}{2x^2 + 4xy^2} \)

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Answer

To show the derivative, we use implicit differentiation on the equation x2y2+xy4=12x^2y^2 + xy^4 = 12.

Differentiating both sides with respect to (x) gives: 2xy2+x(4y3dydx)+2y2dydx=0.2xy^2 + x(4y^3 \frac{dy}{dx}) + 2y^2 \frac{dy}{dx} = 0.

Next, we factor out (\frac{dy}{dx}): x(4y3)dydx+2y2dydx=2xy2.x(4y^3)\frac{dy}{dx} + 2y^2\frac{dy}{dx} = -2xy^2.

This can be rearranged as: dydx(4xy3+2y2)=2xy2.\frac{dy}{dx}(4xy^3 + 2y^2) = -2xy^2.

Now, solving for (\frac{dy}{dx}): dydx=2xy24xy3+2y2.\frac{dy}{dx} = \frac{-2xy^2}{4xy^3 + 2y^2}.

This further simplifies to: dydx=2xy+y32x2+4xy2.\frac{dy}{dx} = \frac{2xy + y^3}{2x^2 + 4xy^2}.

Thus, we have shown the required result.

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