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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.

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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Transformations of quadratic functions

Full solution

Q. 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

.

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

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