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The curve C has equation $$4x^2 - y^3 - 4xy + 2 = 0$$ The point P with coordinates (-2, 4) lies on C - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 5
Question 6
The curve C has equation $$4x^2 - y^3 - 4xy + 2 = 0$$ The point P with coordinates (-2, 4) lies on C. (a) Find the exact value of \( \frac{dy}{dx} \) at the point... show full transcript
Worked Solution & Example Answer:The curve C has equation $$4x^2 - y^3 - 4xy + 2 = 0$$ The point P with coordinates (-2, 4) lies on C - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 5
Step 1
Find the exact value of \( \frac{dy}{dx} \) at the point P.
Answer
To find ( \frac{dy}{dx} ) at the point P (-2, 4), we start by differentiating the equation of the curve implicitly:
-
Differentiate both sides:
[ \frac{d}{dx}(4x^2) - \frac{d}{dx}(y^3) - \frac{d}{dx}(4xy) + \frac{d}{dx}(2) = 0 ] which leads to
[ 8x - 3y^2\frac{dy}{dx} - (4y + 4x\frac{dy}{dx}) = 0 ] -
Rearranging gives us
[ (4y + 4x)\frac{dy}{dx} = 8x - 3y^2 ] -
Solve for ( \frac{dy}{dx} ):
[ \frac{dy}{dx} = \frac{8x - 3y^2}{4y + 4x} ] -
At ( P(-2, 4) ), substitute x = -2 and y = 4:
[ \frac{dy}{dx} = \frac{8(-2) - 3(4)^2}{4(4) + 4(-2)} = \frac{-16 - 48}{16 - 8} = \frac{-64}{8} = -8 ]
Thus, the exact value of ( \frac{dy}{dx} ) at point P is -8.
Step 2
Find the y coordinate of A, giving your answer in the form \( p + q \ln 2 \).
Answer
The normal to the curve at point P forms a straight line. We need to find its slope first:
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The slope of the normal line is the negative reciprocal of ( \frac{dy}{dx} ):
[ m_{normal} = -\frac{1}{-8} = \frac{1}{8} ] -
Using point-slope form, the equation of the normal at P(-2, 4) is: [ y - 4 = \frac{1}{8}(x + 2) ] Rearranging gives
[ y = \frac{1}{8}x + \frac{1}{4} + 4 ]
[ y = \frac{1}{8}x + \frac{17}{4} ] -
To find where this normal meets the y-axis (where ( x = 0 )):
[ y = \frac{17}{4} ] -
Express ( y ) in the required form ( p + q \ln 2 ):
Let ( p = \frac{17}{4} ) and ( q = 0 ). Thus, the y-coordinate of A is:
[ y = \frac{17}{4} + 0 \ln 2 ]