The curve C has equation $y = 2x^3 + kx^2 + 5x + 6$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 1

The gradient of the line given by the equation $2y - 17x - 1 = 0$ can be found by rearranging it into slope-intercept form:

$2y = 17x + 1$ $y = \frac{17}{2}x + \frac{1}{2}$

Thus, the gradient is $\frac{17}{2}$.

At point P where $x = -2$, we can substitute into the derivative found earlier:

$\frac{dy}{dx} = 6(-2)^2 + 2k(-2) + 5$

Calculating $\frac{dy}{dx}$ at $x = -2$ gives:

$\frac{dy}{dx} = 24 - 4k + 5 = 29 - 4k$

Setting this equal to the gradient of the line:

$29 - 4k = \frac{17}{2}$

To solve for $k$, first clear the fraction by multiplying through by 2:

$58 - 8k = 17$

Rearranging gives:

$8k = 58 - 17$ $8k = 41$ $k = \frac{41}{8}$

Substituting $x = -2$ into the original equation to find the y-coordinate of point P:

$y = 2(-2)^3 + k(-2)^2 + 5(-2) + 6$

Now, substituting $k = \frac{41}{8}$ gives:

$y = 2(-8) + \frac{41}{8}(4) - 10 + 6$ $y = -16 + \frac{164}{8} - 10 + 6$ $y = -16 - 10 + 6 + 20.5$ $y = -16 + 20.5 = 4.5$

Thus, the y-coordinate of point P is $4.5$.

The equation of a line in point-slope form is given by:

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

At point P, we have:

• $x_1 = -2$
• $y_1 = 4.5$
• Gradient $m = \frac{17}{2}$.

Plugging in these values:

$y - 4.5 = \frac{17}{2}(x + 2)$

Simplifying this equation:

$y - 4.5 = \frac{17}{2}x + 17$

Which leads to:

$y = \frac{17}{2}x + 21.5$

To write this in the required form $ax + by + c = 0$, we convert:

$-\frac{17}{2}x + y - 21.5 = 0$

Multiplying through by 2 to eliminate the fraction:

$-17x + 2y - 43 = 0$

Thus, the final equation in the desired format is:

$17x - 2y + 43 = 0$