The point P lies on the curve with equation y = 4e^{2x}. The y-coordinate of P is 8. (a) Find, in terms of ln 2, the x-coordinate of P. (b) Find the equation of the tangent to the curve at the point P in the form y = ax + b, where a and b are exact constants to be found.

A-Level Edexcel Maths Pure Solving Equations: The point P lies on the curve wi

The point P lies on the curve with equation y = 4e^{2x} - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 5

Question 2

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The point P lies on the curve with equation y = 4e^{2x}. The y-coordinate of P is 8. (a) Find, in terms of ln 2, the x-coordinate of P. (b) Find the equation of ... show full transcript

Worked Solution & Example Answer:The point P lies on the curve with equation y = 4e^{2x} - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 5

Step 1

Find, in terms of ln 2, the x-coordinate of P.

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Answer

To find the x-coordinate of point P, we start with the given equation of the curve:

$y = 4e^{2x}$

Since we know the y-coordinate is 8:

$8 = 4e^{2x}$

Dividing both sides by 4 gives:

$2 = e^{2x}$

To solve for x, we take the natural logarithm of both sides:

$ext{ln}(2) = 2x$

Dividing by 2 results in:

$x = rac{1}{2} ext{ln}(2)$

Step 2

Find the equation of the tangent to the curve at the point P.

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Answer

To find the equation of the tangent at point P, we first need to calculate the derivative of y with respect to x:

Differentiate y:

$rac{dy}{dx} = 8e^{2x}$
Substitute the x-coordinate we found:

$x = rac{1}{2} ext{ln}(2)$

Thus,

$rac{dy}{dx} = 8e^{2 imes rac{1}{2} ext{ln}(2)} = 8e^{ ext{ln}(2)} = 8 imes 2 = 16$
Now, using the equation of the tangent line in the form y = ax + b:

Using the point-slope form, where slope a = 16 and the point P is (x, 8):

$y - 8 = 16(x - rac{1}{2} ext{ln}(2))$

Rearranging gives:

$y - 8 = 16x - 8 ext{ln}(2)$

This simplifies to:

$y = 16x + (8 - 8 ext{ln}(2))$

The point P lies on the curve with equation y = 4e^{2x} - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 5

Worked Solution & Example Answer:The point P lies on the curve with equation y = 4e^{2x} - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 5

Find, in terms of ln 2, the x-coordinate of P.

Find the equation of the tangent to the curve at the point P.

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