Maximium and minimum value of area.

Question

Given that the equation of parabola is $y=x^2+1,1\leq x\leq 3,$ what is the maximum and minimum value of area formed by $x$-axis, tangent, normal at any point on parabola? Now I wrote the equation as $x^2=4\times 0.25(y-1)$ so focus is at $0,1.$ Equation of normal, tangent for a point on parabola is $(y-y_1)=\frac{-y_1}{2a}(x-x_1),xx_1=2a(y+y_1)$ respectively. Now area of triangle is $0.5\times b\times h$ so I need to get $b,h$ ie base, height as function of $x$ and then differentiate but I am struggling now hope you guys help. Thanks

Answer 1

Assume a point $(u,v)$ on the parabola, because the derivative of function $y=x^2+1$ is $y' = 2x$.

Therefore the tangent line passing through $(u,v)$ has equation: $y - v = 2u(x-u)$...(1) (i.e. statement in the text: ($xx_1=2a(y+y_1)$) )

In the same way, the norm line passing through $(u,v)$ has equation: $ y -v = - 1/(2u) (x-u)$...(2)

Setting $y = 0$ in equation $(1)$, $ x_1 = \frac{-v}{2u}+u $

Setting $y = 0$ in equation $(2)$, $ x_2 = 2uv+u$

As the text above mentioned, the area is $1/2 \times (x_2-x_1) \times v = 1/2 \times (2uv^2+v^2/(2u)) = 1/2 \times(2u(u^2+1)^2+(u^2+1)^2/(2u)) $

The derivative of area with respect to u is $10u^4+27u^2/2-1/(2u^3)+3$

Find the stationary point(s):let this be zero: => $20u^4+7u^2 = 1$

=> $u = 1/2 \sqrt{\frac{\sqrt{129}}{10} - 7/10}$ (the negative value is ignored ). And this is not included in the domain.

Therefore, the area as a function with respect to u is monotonously increasing.

The area is minimal when u = 1, which equals to 5

The area is maximal when u = 3, which equals to $925/3$