# Maximium and minimum value of area.

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

• The derivative of $y= x^2+ 1$ at x= a is $y'= 2a$. The tangent line at that point is $y= 2a(x- a)+ a^2+ 1$. The normal line at that point is $y= -\frac{1}{2a}(x- a)+ a^2+ 1$. The tangent line cuts the x-axis where $2a(x- a)+ a^2+ 1= 0$ so at $((\frac{a}{2}- \frac{1}{2a}, 0)$. The normal line cuts the x-axis where $-\frac{1}{2a}(x- a)+ a^2+ 1= 0$ so at $(2a^3+ a 0)$. The area of a right triangle is 1/2 the product of the lengths of the legs so find the distance from each of those two points to $(a, a^2+ 1)$. Jun 5 '16 at 13:27

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 $925/3$

• Please explain me what you have done after derivative Jun 5 '16 at 13:46
• Find the stationary point found that the stationary point's horizontal coordinate is not within [1,3] Jun 5 '16 at 13:48
• So it shows that the area is increasing over the interval [1,3] Jun 5 '16 at 13:48
• No not montonicity ,the simplification of polynomial Jun 5 '16 at 13:54
• $10u^4+27u^2/2-1/(2u^2)+3 = (u^2+1)(20u^4+7u^2-1)/2u^2 = 0$ Jun 5 '16 at 13:56