A-Level Edexcel Maths Pure Vectors in 3 Dimensions: 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 6 - 2008 - Paper 8

Question 6

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 6-2008-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 6 - 2008 - Paper 8

Step 1

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

96%

114 rated

Answer

To find the coordinates of the points on the curve where $x = -8$ , substitute $x$ into the equation:

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

This simplifies to:

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

Rearranging gives:

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

Dividing through by 4, we have:

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

Next, we can apply the quadratic formula where $a = 1$ , $b = -24$ , and $c = -128$ :

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{24 \pm \sqrt{576 + 512}}{2}$

Calculating the discriminant:

$\sqrt{1088} = \sqrt{16 \times 68} = 4\sqrt{68} \approx 4 \times 8.2462 = 32.984$

Thus, we find:

$y \approx \frac{24 \pm 32.984}{2}$

Calculating the two values:

For the positive root:

$y \approx \frac{56.984}{2} = 28.492$

For the negative root:

$y \approx \frac{-8.984}{2} = -4.492$

Thus, the two points are approximately $(-8, 28.492)$ and $(-8, -4.492)$ .

Step 2

Find the gradient of the curve at each of these points.

99%

104 rated

Answer

To find the gradient of the curve, we need to differentiate the equation:

$\frac{dy}{dx}$

Differentiate both sides implicitly:

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

Rearranging gives:

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

Then solve for $rac{dy}{dx}$ :

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

Now, substituting $x = -8$ and $y = 28.492$ :

$\frac{dy}{dx} = \frac{3(-8)^2 - 12(28.492)}{12(-8) + 8(28.492)}$

Calculating yields:

$\frac{dy}{dx} = \frac{192 - 341.904}{-96 + 227.936} = \frac{-149.904}{131.936} \approx -1.1348$

Next, substituting $x = -8$ and $y = -4.492$ :

$\frac{dy}{dx} = \frac{192 - (-53.904)}{-96 + (-35.936)} = \frac{245.904}{-131.936} \approx -1.8608$

Thus, the gradients at the points are approximately $-1.1348$ and $-1.8608$ .

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 6 - 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 6 - 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.

Join the A-Level students using SimpleStudy...