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

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

Worked Solution & Example Answer:A curve is described by the equation $$x^2 - 4y^2 = 12xy.$$ (a) Find the coordinates of the two points on the curve where x = -8 - 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, substitute x = -8 into the equation:

(−8)2−4y2=12(−8)y.(-8)^2 - 4y^2 = 12(-8)y.
This simplifies to:

64−4y2=−96y.64 - 4y^2 = -96y.
Rearranging gives:

4y2−96y+64=0.4y^2 - 96y + 64 = 0.
Dividing the whole equation by 4 results in:

y2−24y+16=0.y^2 - 24y + 16 = 0.
Now we will apply the quadratic formula:

y=−b±b2−4ac2a=24±576−642=24±5122=12±5122.y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{24 \pm \sqrt{576 - 64}}{2} = \frac{24 \pm \sqrt{512}}{2} = 12 \pm \frac{\sqrt{512}}{2}.
Calculating further, we find:

512=162,\sqrt{512} = 16\sqrt{2},
Thus:

y=12±82.y = 12 \pm 8\sqrt{2}.
This leads to the coordinates:

  1. (−8,12+82)(-8, 12 + 8\sqrt{2})
  2. (−8,12−82)(-8, 12 - 8\sqrt{2}).

Step 2

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

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Answer

To find the gradient, we need to differentiate the equation implicitly:

dydx=3x2−12y12y+8x. \frac{dy}{dx} = \frac{3x^2 - 12y}{12y + 8x}.
Substituting x = -8 into the gradient formula:

At the first point:

  1. For y=12+82y = 12 + 8\sqrt{2}:

    • Substitute: dydx=3(−8)2−12(12+82)12(12+82)+8(−8). \frac{dy}{dx} = \frac{3(-8)^2 - 12(12 + 8\sqrt{2})}{12(12 + 8\sqrt{2}) + 8(-8)}.
    • This simplifies as: =192−144−962144+962−64=48−96280+962. = \frac{192 - 144 - 96\sqrt{2}}{144 + 96\sqrt{2} - 64} = \frac{48 - 96\sqrt{2}}{80 + 96\sqrt{2}}.

  2. For y=12−82y = 12 - 8\sqrt{2}:

    • Substitute: dydx=192−144+962144−962−64=48+96280−962. \frac{dy}{dx} = \frac{192 - 144 + 96\sqrt{2}}{144 - 96\sqrt{2} - 64} = \frac{48 + 96\sqrt{2}}{80 - 96\sqrt{2}}.
      Both values provide the gradients at each corresponding point.

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