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

Explanation:

Step1: Determine the vertical distance from the centre of the dish

The vertical distance from the wall is 30.6 cm. Since the parabola is symmetric about its axis, and assuming the wall - dish connection is at the edge of the parabolic cross - section, we need to find the value of \(y\) at a certain \(x\). First, we note that the equation of the parabola is \(y=\frac{1}{15}x^{2}+4x\). The vertical distance from the centre of the dish is the value of \(y\) at the relevant \(x\) value. We know that the parabola is symmetric, and we need to find the \(x\) value corresponding to the given vertical position. However, we can also use the fact that if we assume the centre of the dish is at \(x = 0\), and we want to find the horizontal distance \(x\) when \(y\) is related to the given vertical distance from the wall. Let's assume the vertex of the parabola is at the centre of the dish. We are given the vertical distance from the wall. We need to set \(y\) to the appropriate value and solve for \(x\).
The vertical distance from the centre of the dish to the wall is related to the \(y\) value in the parabola equation. If we assume the centre of the dish is at \(y = 0\) (for the purpose of calculating the horizontal displacement), and the vertical distance from the wall to the centre of the dish is \(y=30.6\) cm. We set \(y = 30.6\) in the equation \(y=\frac{1}{15}x^{2}+4x\). So we have the quadratic equation \(\frac{1}{15}x^{2}+4x - 30.6=0\). Multiply through by 15 to get \(x^{2}+60x - 459 = 0\).

Step2: Solve the quadratic equation

The quadratic formula for \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 1\), \(b = 60\) 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 \(x\) represents a distance, so \(x=\frac{-60 + 73.73}{2}=\frac{13.73}{2}\approx6\) cm.

Answer:

d) 6 cm