A curve is such that (dy)/(dx)=(6)/((2x-1)^(2)) and P(2,9) is a point on the curve. The normal to the curve at P meets the y-axis at Q and the x-axis at R. Find the coordinates of the midpoint of QR.

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Differentiate and Identify Gradient: Identify the gradient of the curve at any point by differentiating the given equation. Given: $\frac{dy}{dx} = \frac{6}{(2x - 1)^2}$Find Normal Gradient at P(2,9): Find the gradient of the normal to the curve at point P(2,9). The gradient of the normal is the negative reciprocal of the gradient of the curve. Gradient of the curve at P: $\frac{dy}{dx} = \frac{6}{(2*2 - 1)^2} = \frac{6}{(3)^2} = \frac{6}{9} = \frac{2}{3}$ Gradient of the normal at P: $-\frac{3}{2}$Equation of Normal through P: Use the point-gradient form to find the equation of the normal. Point-gradient form: $y - y_1 = m(x - x_1)$ Using point P(2,9) and gradient $-\frac{3}{2}$, we get: $y - 9 = -\frac{3}{2}(x - 2)$Find Y-Intercept Q: Rearrange the equation of the normal to find the y-intercept (Q). $y = -\frac{3}{2}x + 3 + 9$ $y = -\frac{3}{2}x + 12$ The y-intercept Q is at (0, 12).Find X-Intercept R: Find the x-intercept (R) by setting y to 0 in the equation of the normal. $0 = -\frac{3}{2}x + 12$ $\frac{3}{2}x = 12$ $x = \frac{12}{\frac{3}{2}}$ $x = \frac{12 * 2}{3}$ $x = 8$ The x-intercept R is at (8, 0).Calculate Midpoint of QR: Calculate the midpoint of QR using the midpoint formula. Midpoint formula: $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ Using Q(0, 12) and R(8, 0), we get: Midpoint of QR: $\left(\frac{0 + 8}{2}, \frac{12 + 0}{2}\right)$ Midpoint of QR: $(4, 6)$

A curve is such that $\frac{d y}{d x}=\frac{6}{(2 x-1)^{2}}$ and $P(2,9)$ is a point on the curve. The normal to the curve at $P$ meets the $y$ -axis at $Q$ and the $x$ -axts at $R$ . Find the coordinates of the midpoint of $Q R$ .

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A curve is such that dydx=6(2x−1)2 \frac{d y}{d x}=\frac{6}{(2 x-1)^{2}} dxdy​=(2x−1)26​ and P(2,9) P(2,9) P(2,9) is a point on the curve. The normal to the curve at P P P meets the y y y-axis at Q Q Q and the x x x-axts at R R R. Find the coordinates of the midpoint of QR Q R QR.

A curve is such that $\frac{d y}{d x}=\frac{6}{(2 x-1)^{2}}$ and $P(2,9)$ is a point on the curve. The normal to the curve at $P$ meets the $y$ -axis at $Q$ and the $x$ -axts at $R$ . Find the coordinates of the midpoint of $Q R$ .