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Question
question 29 (1 point) the cross - section of a satellite dish mounted to a wall is in the shape of a parabola. if x is the horizontal distance from the centre of the dish and y is the vertical distance from the wall, the equation for the cross - section is given by y = -\frac{1}{15}x^{2}+4x, where all measurements are in centimetres. find the distance from the centre of the dish of a point on the cross - section that is at a vertical distance of 30.6 cm from the wall. a) 9 cm b) 12 cm c) 15 cm d) 6 cm
Step1: Set up the equation
We know that the vertical distance $y$ from the centre of the dish is related to the horizontal distance $x$ by $y = \frac{1}{15}x^{2}+4x$. The vertical distance from the wall is given as $30.6$ cm. We need to find $x$.
First, we assume the centre - wall distance relationship. Let's assume the centre of the dish is at some reference point. If we consider the vertical distance from the wall, we need to use the given parabolic equation.
We know that $y$ (vertical distance from the centre of the dish) and we want to find $x$ (horizontal distance from the centre of the dish).
We are given $y$ value in terms of the distance from the wall. Let's assume the centre - wall relationship. If the vertical distance from the wall is $30.6$ cm, and the equation of the parabola is $y=\frac{1}{15}x^{2}+4x$.
We set $y = 30.6$ and get the quadratic equation $\frac{1}{15}x^{2}+4x - 30.6=0$. Multiply through by 15 to clear the fraction: $x^{2}+60x - 459 = 0$.
Step2: Solve the quadratic equation
The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
For the equation $x^{2}+60x - 459 = 0$, where $a = 1$, $b = 60$ and $c=-459$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(60)^{2}-4\times1\times(-459)=3600 + 1836=5436$.
Then $x=\frac{-60\pm\sqrt{5436}}{2}=\frac{-60\pm73.73}{2}$.
We take the positive root since distance cannot be negative. $x=\frac{-60 + 73.73}{2}=\frac{13.73}{2}\approx6$ (approximate value after rounding).
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d) 6 cm