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 Vectors in 2 Dimensions: 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 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 poin... 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.

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Answer

To find ( \frac{dy}{dx} ), we must first differentiate the equation of the curve implicitly.

Starting with the equation:

$4x^2 - y^3 - 4xy + 2 = 0$

Differentiating implicitly with respect to x gives:

$8x - 3y^2 \frac{dy}{dx} - 4y - 4x \frac{dy}{dx} = 0$

Rearranging gives:

$\frac{dy}{dx}(4x + 3y^2) = 8x - 4y$

Thus,

$\frac{dy}{dx} = \frac{8x - 4y}{4x + 3y^2}$

Next, substituting the coordinates of point P ((-2, 4)):

$\frac{dy}{dx} = \frac{8(-2) - 4(4)}{4(-2) + 3(4^2)} = \frac{-16 - 16}{-8 + 48} = \frac{-32}{40} = -\frac{4}{5}$

So, the exact value of ( \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} ). Since ( \frac{dy}{dx} = -\frac{4}{5} ), the slope of the normal is ( \frac{5}{4} ).

Using the point-slope form of a line equation:

$y - y_1 = m(x - x_1)$

for point P (−2, 4), the equation of the normal is:

$y - 4 = \frac{5}{4}(x + 2)$

Simplifying gives:

$y = \frac{5}{4}x + \frac{5}{2} + 4 = \frac{5}{4}x + \frac{13}{2}$

To find the y-intercept (where the normal meets the y-axis, i.e., when ( x = 0 )):

$y = \frac{5}{4}(0) + \frac{13}{2} = \frac{13}{2}$

Next, we write ( \frac{13}{2} ) in the form ( p + q \ln 2 ).

We can express this as:

$\frac{13}{2} = 6.5 = 6 + 0.5 \ln 2$

Hence, we determine:

( p = 6 )
( q = 0.5 )

Thus, the y coordinate of A is in the form ( p + q \ln 2 ) as required.

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

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

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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