Question

eqn of normal \(y=m x-2 m-m^{3}\) It passes through \((6,0)\) \(\mathrm{m}^{3}-4 \mathrm{~m}=0\) \(\mathrm{m}\left(\mathrm{m}^{2}-4\right)=0\) \(\mathrm{m}=0, \pm 2\) Pts are \(\left(a m^{2}-2 a m\right)\) \((0,0),(4,-4),(4,4)\) Minimum distance \(=\sqrt{21}-\sqrt{5}\)

Short answer

Answer: The minimum distance is \(6 - 2\sqrt{5}\) units.

Step-by-step solution

1. Analyze the given information

We are given the following pieces of information: - Equation of the normal: \(y=mx-2m-m^3\) - It passes through the point \((6,0)\) - The polynomial equation related to the slope of the normal: \(m^3 - 4m = 0\) The goal is to find the points where the normal passes through and compute the minimum distance.

2. Solve the polynomial equation for slope m

We are given the equation: \(m^3 - 4m = 0\). By factoring, we have: \(m(m^2-4) = 0\) This implies that \(m=0\), \(m=+2\), or \(m=-2\).

3. Plug the values of m into the equation of the normal

We have three possible values of the slope (0, 2, and -2). We can plug each value into the equation of the normal to find the points it passes through. When \(m=0\): \(y=0\cdot x - 2\cdot0 - 0^3 = 0\) This is the horizontal line and intersects the x-axis. When \(m=2\): \(y=2x - 2(2) - 2^3 = 2x - 16\) This line passes through \((4, -4)\). When \(m=-2\): \(y=-2x - 2(-2) - (-2)^3 = -2x + 16\) This line passes through \((4, 4)\).

4. Calculate the minimum distance from the point (6,0)

Now, we have three points that the normal passes through: \((0,0)\), \((4,-4)\), and \((4,4)\). We will find the distances from these points to the given point \((6,0)\). • Distance from \((0,0)\) to \((6,0)\): \(D_1 = \sqrt{(6-0)^2 + (0-0)^2} = 6\) • Distance from \((4,-4)\) to \((6,0)\): \(D_2 = \sqrt{(6-4)^2 + (0+4)^2} = \sqrt{4 + 16} = \sqrt{20}\) • Distance from \((4,4)\) to \((6,0)\): \(D_3 = \sqrt{(6-4)^2 + (0-4)^2} = \sqrt{4 + 16} = \sqrt{20}\) Finally, the minimum distance is the smallest among these distances, which is \(D_1\): Minimum Distance = \(6 - 2\sqrt{5}\)