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

  • $\begingroup$ 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 $(a, a^2+ 1)$. $\endgroup$
    – user247327
    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 $925/3$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.