Question

\(y=\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}-4-\frac{x_{2}^{2}}{4}\right)^{2}\) Min value of \(y\) is shortest distance between \(y=x-4\) \& \(y=\frac{x^{2}}{4}\) slope of normal \(=-1\) \(\frac{d y}{d x}=\frac{x}{2}\) \(-\frac{d x}{d y}=\frac{-2}{x}=-1\) \(\mathrm{x}=2\)

Short answer

Answer: The minimum value of the function is 20.

Step-by-step solution

Find the Intersection Points

First, we need to find the points where these two curves intersect. To do this, set \(x-4=\frac{x^{2}}{4}\).

Solve for \(x\)

Rearranging the equation from step 1 and solving for \(x\): $$x^{2}-4x+16=0$$ Let's solve the quadratic equation to find the possible values of \(x\): $$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{4\pm\sqrt{(-4)^{2}-4*16}}{2}=\frac{4\pm\sqrt{0}}{2}=2$$

Calculate the Corresponding \(y\)

To find the corresponding value of \(y\), plug \(x=2\) into either of the given equations. We'll use \(y=x-4\): $$y=2-4=-2$$

Verify the Slope of the Normal

Now, we need to verify whether the given slope of the normal is \(-1\) for the intersection point \((2,-2)\). Use the given equations of \(\frac{dy}{dx}=\frac{x}{2}\) and \(-\frac{dx}{dy}=\frac{-2}{x}=-1\): $$\frac{dy}{dx}=\frac{2}{2}=1$$ $$-\frac{dx}{dy}=\frac{-2}{2}=-1$$ Yes, the slope of the normal is \(-1\).

Calculate the Minimum Value of the Function

At the intersection point \((2,-2)\), plug in these values for \(x_{1}\) and \(x_{2}\) into the function \(y=\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}-4-\frac{x_{2}^{2}}{4}\right)^{2}\): $$y=\left(2-(-2)\right)^{2}+\left(2-4-\frac{(-2)^{2}}{4}\right)^{2}=(4)^{2}+(-2)^{2}=16+4=20$$ The minimum value of the given function \(y\) is \(20\).