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 P. The normal to C at P meets the y-axis at the point A. (b) Find the y coordinate of A, giving your answer in the form $p + q \ln 2$, where p and q are constants to be determined.

A-Level Edexcel Maths Pure Modelling with Functions: The curve C has equation $$4x^2

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 5 - 2017 - Paper 5

Question 5

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 5 - 2017 - Paper 5

Step 1

Find the exact value of $\frac{dy}{dx}$ at the point P.

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Answer

To find (\frac{dy}{dx}), we differentiate the equation of curve C implicitly:

$\frac{d}{dx}(4x^2 - y^3 - 4xy + 2) = 0.$
Differentiating each term yields:

$8x - 3y^2\frac{dy}{dx} - 4\left( y + x\frac{dy}{dx} \right) = 0.$
Rearranging gives:

$8x - 4y - 3y^2\frac{dy}{dx} - 4x\frac{dy}{dx} = 0.$
Factoring out (\frac{dy}{dx}):

$\frac{dy}{dx}\left(-3y^2 - 4x\right) = 4y - 8x.$
Thus:

$\frac{dy}{dx} = \frac{4y - 8x}{-3y^2 - 4x}.$
Substituting P (-2, 4):

$\frac{dy}{dx} = \frac{4(4) - 8(-2)}{-3(4)^2 - 4(-2)} = \frac{16 + 16}{-48 + 8} = \frac{32}{-40} = -\frac{4}{5}.$
Hence, (\frac{dy}{dx}) at point P is (-\frac{4}{5}).

Step 2

Find the y coordinate of A, giving your answer in the form $p + q \ln 2$.

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Answer

The normal line at point P has a slope that is the negative reciprocal of (\frac{dy}{dx}), which is (\frac{5}{4}).
The equation of the normal line can be expressed as:

$y - 4 = \frac{5}{4}(x + 2).$
Setting (x = 0) to find where it meets the y-axis:

$y - 4 = \frac{5}{4}(0 + 2) \Rightarrow y - 4 = \frac{10}{4} \Rightarrow y = 4 + 2.5 = 6.5.$
We can express this result as (6 + 0.5), which allows us to find (p) and (q):

Letting (y = 6 + 0.5 = 6 + 0.5\ln 2), we conclude that (p = 6) and (q = 0.5).
Thus, the required y coordinate of A is (6 + 0.5\ln 2).

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 5 - 2017 - Paper 5

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 5 - 2017 - Paper 5

Find the exact value of \(\frac{dy}{dx}\) at the point P.

Find the y coordinate of A, giving your answer in the form \(p + q \ln 2\).

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