What is the equation of the line with the property that the shortest distance between that line and the parabola is 1?
The parabola is $y=x^2$ and the linear pass (2,0). The shortest distance between that linear and the parabola is 1.
Normally, linear & parabola calculus problem will ask the equation of linear that touch the parabola at specific (x,y) point. I can solve that problem by finding the derivative of the parabola. Then put the x value into the derivative function to find the slope of the linear. After that, it isn't hard to find the equation of the linear.
But this question is different. It ask the equation of linear that have the shortest distance between that linear and the parabola is 1. I don't know how solve it. I know that the linear y=ax-2a since it pass (2,0) but don't know the next step.
I try plotting graph in Desmos. I think $a \approx 6\pm1$ but don't know the exact value.
Best Answer
At the point $(x_0,x_0^2)$ the parabola $y=x^2$ has tangent line $y = 2x_0(x-x_0) + x_0^2 = 2x_0x - x_0^2$.
If we slide over the line $y = ax-2a$ by the right amount, this line should become a tangent line. To find the right amount to make the normal distance equal to one, use the perpendicular slope of $-\frac{1}{a}$ and draw a right triangle with side 1, which should give you $\sqrt{\frac{a^2+1}{a^2}}$. Then the translated line has formula $y = ax - (2a - a\sqrt{\frac{a^2+1}{a^2}})$
That means we have two equations for the same tangent line, which yields $a = 2x_0$ and $2a - a \sqrt{\frac{a^2+1}{a^2}} = x_0^2$.
That gives a quartic for $a$, and WolframAlpha gives $a = 3.8685$ or $a = 0.64851$, or the degenerate $a = -0.53086$ or $a = 12.014$, which should get you the equation for the line.