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leprofece
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the normal at a point N of the parabole: y2 = 2px to the curve at another point M. calculate the intercept of N when the length of MN is minimal.
Answer sqrt (2) p
Answer sqrt (2) p
MarkFL said:I would begin be letting point $N$ be \(\displaystyle \left(\frac{y^2}{2p},y\right)\) and point $M$ be \(\displaystyle \left(\frac{u^2}{2p},u\right)\). Now, you know what the slope of the line through $M$ and $N$ has to be since this line is normal to the given parabola at $N$, so this will allow you to express $u$ as a function of $y$ and the parameter $p$.
Then you can construct a function representing the distance (or the square of the distance) between $M$ and $N$ which you can minimize. Once you have the particular $y$ critical $y$-value you can then find the root or $x$-intercept of this line.
MarkFL said:You are correct in that the slope of the line is:
\(\displaystyle m=\frac{2p}{y+u}\)
(I simplified a bit).
Now, can you use the calculus to come up with another expression for the slope using the fact that the line is normal to the parabola at $N$?
MarkFL said:You have:
\(\displaystyle x=\frac{y^2}{2p}\)
So, what is \(\displaystyle -\frac{dx}{dy}\)?
MarkFL said:Correct, but you want the negative of this. And now you have two expressions representing the slope of the normal line and you may equate them and get $u$ as a function of $y$. Then continue following the instructions I gave in my first post. :D
MarkFL said:I get a different value for $u$...how did you obtain the value you have?
MarkFL said:I got:
\(\displaystyle u=-\frac{2p^2+y^2}{y}\)
And so the objective function (the square of $\overline{MN}$) is:
\(\displaystyle f(y)=\left(\frac{y^2}{2p}-\frac{\dfrac{2p^2+y^2}{y}}{2p}\right)^2+\left(y+\frac{2p^2+y^2}{y}\right)^2\)
After some algebraic simplification, I obtained:
\(\displaystyle f(y)=4\left(\left(\frac{p}{y}\right)^2+1\right)\left(y^2+1\right)\)
Now, I will leave you to verify this, and then minimize the objective function.
leprofece said:Thanks now I derived this expression and got
8(1+p2)(y-y2+p2)/(y2)
and I equated to 0
and got y2-y-p2 but I didnt get anything
leprofece said:I have just derived the expression that we have got before
A max and min parabola is a type of curved graph that represents a quadratic function. It has a single highest point (maximum) or lowest point (minimum) on the curve, known as the vertex.
The vertex of a max and min parabola can be found by using the formula x = -b/2a, where a and b are the coefficients of the quadratic equation. This gives the x-coordinate of the vertex, and the y-coordinate can be found by substituting the x-value into the original equation.
A maximum on a parabola is the highest point on the curve, while a minimum is the lowest point. Mathematically, a maximum has a positive coefficient for the squared term (a>0), while a minimum has a negative coefficient (a<0).
No, a max and min parabola can only have one maximum or minimum. This is because the graph of a quadratic function is a parabola, which has a single curve and therefore can only have one highest or lowest point.
Max and min parabolas have many real-life applications, such as in physics to model the trajectory of a projectile, in engineering to design bridges and buildings, and in economics to analyze profit and cost functions. They can also be used in data analysis to identify trends and make predictions.