A-Level Edexcel Maths Pure Integration: A curve is described by the equa

A curve is described by the equation $x^3 - 4y^2 = 12xy.$ (a) Find the coordinates of the two points on the curve where $x = -8.$ (b) Find the gradient of the curve at each of these points. - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 8

Question 7

A-curve-is-described-by-the-equation--$x^3---4y^2-=-12xy.$--(a)-Find-the-coordinates-of-the-two-points-on-the-curve-where-$x-=--8.$--(b)-Find-the-gradient-of-the-curve-at-each-of-these-points.-Edexcel-A-Level Maths Pure-Question 7-2008-Paper 8.png

A curve is described by the equation $x^3 - 4y^2 = 12xy.$ (a) Find the coordinates of the two points on the curve where $x = -8.$ (b) Find the gradient of the cur... show full transcript

Worked Solution & Example Answer:A curve is described by the equation $x^3 - 4y^2 = 12xy.$ (a) Find the coordinates of the two points on the curve where $x = -8.$ (b) Find the gradient of the curve at each of these points. - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 8

Step 1

Find the coordinates of the two points on the curve where $x = -8$

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Answer

To find the coordinates, we substitute $x = -8$ into the equation:

$(-8)^3 - 4y^2 = 12(-8)y$

This simplifies to:

$-512 - 4y^2 = -96y$

Rearranging gives:

$4y^2 - 96y - 512 = 0$

Now, we can simplify this by dividing through by 4:

$y^2 - 24y - 128 = 0$

Next, we use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$ , $b = -24$ , and $c = -128$ . This leads us to:

$y = \frac{24 \pm \sqrt{(-24)^2 - 4(1)(-128)}}{2(1)}$

Calculating the discriminant:

$(-24)^2 - 4(1)(-128) = 576 + 512 = 1088$

Thus, we have:

$y = \frac{24 \pm \sqrt{1088}}{2} = \frac{24 \pm 32.96}{2}$

Calculating both possible values for $y$ :

$y = \frac{24 + 32.96}{2} = 28.48$ \
$y = \frac{24 - 32.96}{2} = -4.48$

Thus, the coordinates are approximately $(-8, 28.48)$ and $(-8, -4.48)$ .

Step 2

Find the gradient of the curve at each of these points

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Answer

To find the gradient of the curve, we must differentiate the equation implicitly:

$\frac{dy}{dx}$ . Differentiating both sides gives:

$3x^2 - 8y\frac{dy}{dx} = 12y + 12x\frac{dy}{dx}$

Rearranging, we get:

$(3x^2 - 12y) = (12x + 8y)\frac{dy}{dx}$

Thus:

$\frac{dy}{dx} = \frac{3x^2 - 12y}{12x + 8y}$

Substituting $x = -8$ and $y = 28.48$ :

$\frac{dy}{dx} = \frac{3(-8)^2 - 12(28.48)}{12(-8) + 8(28.48)}$

Calculating:

$= \frac{192 - 341.76}{-96 + 227.84} = \frac{-149.76}{131.84} \approx -1.13$

Now substituting $y = -4.48$ :

$\frac{dy}{dx} = \frac{192 - (12)(-4.48)}{-96 + 8(-4.48)}$

Calculating this yields:

$= \frac{192 + 53.76}{-96 - 35.84} = \frac{245.76}{-131.84} \approx -1.86$

Therefore, the gradients at the points are approximately -1.13 and -1.86.

A curve is described by the equation $x^3 - 4y^2 = 12xy.$ (a) Find the coordinates of the two points on the curve where $x = -8.$ (b) Find the gradient of the curve at each of these points. - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 8

Worked Solution & Example Answer:A curve is described by the equation $x^3 - 4y^2 = 12xy.$ (a) Find the coordinates of the two points on the curve where $x = -8.$ (b) Find the gradient of the curve at each of these points. - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 8

Find the coordinates of the two points on the curve where $x = -8$

Find the gradient of the curve at each of these points

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