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The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find $\frac{dy}{dx}$ in terms of x and y - Edexcel - A-Level Maths: Pure - Question 3 - 2015 - Paper 4

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The-curve-C-has-equation--$$x^2---3xy---4y^2-+-64-=-0$$--(a)-Find-$\frac{dy}{dx}$-in-terms-of-x-and-y-Edexcel-A-Level Maths: Pure-Question 3-2015-Paper 4.png

The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find $\frac{dy}{dx}$ in terms of x and y. (b) Find the coordinates of the points on C where $\frac{dy}{... show full transcript

Worked Solution & Example Answer:The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find $\frac{dy}{dx}$ in terms of x and y - Edexcel - A-Level Maths: Pure - Question 3 - 2015 - Paper 4

Step 1

Find $\frac{dy}{dx}$ in terms of x and y.

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Answer

To find dydx\frac{dy}{dx}, we will perform implicit differentiation on the equation of the curve:

Starting from:

x2−3xy−4y2+64=0x^2 - 3xy - 4y^2 + 64 = 0

Differentiating both sides with respect to x:

ddx(x2)−3(xdydx+y)−4⋅2ydydx=0\frac{d}{dx}(x^2) - 3\left( x \frac{dy}{dx} + y \right) - 4 \cdot 2y \frac{dy}{dx} = 0

This leads to:

2x−3(xdydx+y)−8ydydx=02x - 3 \left( x \frac{dy}{dx} + y \right) - 8y \frac{dy}{dx} = 0

Expanding and collecting terms gives:

2x−3xdydx−3y−8ydydx=02x - 3x \frac{dy}{dx} - 3y - 8y \frac{dy}{dx} = 0

Rearranging the equation:

dydx(−3x−8y)=3y−2x\frac{dy}{dx}( -3x - 8y) = 3y - 2x

Thus, we can solve for dydx\frac{dy}{dx}:

dydx=3y−2x−3x−8y\frac{dy}{dx} = \frac{3y - 2x}{-3x - 8y}

Step 2

Find the coordinates of the points on C where $\frac{dy}{dx} = 0$.

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Answer

Setting the numerator of dydx\frac{dy}{dx} equal to zero gives:

3y−2x=03y - 2x = 0

From here, we can express y in terms of x:

y=23xy = \frac{2}{3}x

We will substitute this into the original equation to find the coordinates:

Substituting y=23xy = \frac{2}{3}x into:

x2−3xy−4y2+64=0x^2 - 3xy - 4y^2 + 64 = 0

We have:

x2−3x(23x)−4(23x)2+64=0x^2 - 3x\left(\frac{2}{3}x\right) - 4\left(\frac{2}{3}x\right)^2 + 64 = 0

Simplifying this:

x2−2x2−169x2+64=0x^2 - 2x^2 - \frac{16}{9}x^2 + 64 = 0

This gives:

(1−2−169)x2+64=0\left(1 - 2 - \frac{16}{9}\right)x^2 + 64 = 0

Combining the x terms:

(−9−169)x2+64=0\left(\frac{-9 - 16}{9}\right)x^2 + 64 = 0

(−259)x2+64=0\left(-\frac{25}{9}\right)x^2 + 64 = 0

Solving for x gives:

x2=64⋅925=57625  ⟹  x=±245x^2 = \frac{64 \cdot 9}{25} = \frac{576}{25}\implies x = \pm \frac{24}{5}

Now substituting back to find y:

When x=245x = \frac{24}{5}:

y=23(245)=4815=165y = \frac{2}{3}\left(\frac{24}{5}\right) = \frac{48}{15} = \frac{16}{5}

When x=−245x = -\frac{24}{5}:

y=23(−245)=−4815=−165y = \frac{2}{3}\left(-\frac{24}{5}\right) = -\frac{48}{15} = -\frac{16}{5}

Thus, the points are:

(245,165)and(−245,−165)\left(\frac{24}{5}, \frac{16}{5}\right)\quad \text{and} \quad \left(-\frac{24}{5}, -\frac{16}{5}\right)

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