Question

question 26 (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) 6 cm b) 9 cm c) 12 cm d) 15 cm

Explanation:

Step1: Substitute y - value

We are given the equation $y =-\frac{1}{15}x^{2}+4x$ and $y = 30.6$. So we set up the equation $30.6=-\frac{1}{15}x^{2}+4x$. Multiply through by 15 to clear the fraction: $459=-x^{2}+60x$. Rearrange it to the standard - quadratic form $x^{2}-60x + 459 = 0$.

Step2: Use the quadratic formula

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 $x^{2}-60x + 459 = 0$, we have $a = 1$, $b=-60$, and $c = 459$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-60)^{2}-4\times1\times459=3600 - 1836 = 1764$. Then, $x=\frac{60\pm\sqrt{1764}}{2}=\frac{60\pm42}{2}$.

Step3: Find the two solutions for x

We have two solutions: $x_1=\frac{60 + 42}{2}=\frac{102}{2}=51$ and $x_2=\frac{60 - 42}{2}=\frac{18}{2}=9$. Since $x$ represents a distance, we take the non - negative value that makes physical sense in the context of the problem.

Answer:

B. 9 cm