where does the normal line to the parabola, given below, at the given point, intersect the parabola a second…

Explanation:

Step1: Find the derivative of the parabola

Differentiate $y = 25x-5x^{2}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $y'=25 - 10x$.

Step2: Find the slope of the tangent at the point $P=(5,0)$

Substitute $x = 5$ into $y'$. So $y'(5)=25-10\times5=25 - 50=-25$.

Step3: Find the slope of the normal line

The slope of the normal line $m_n$ is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent at $(5,0)$ is $- 25$, the slope of the normal line $m_n=\frac{1}{25}$.

Step4: Find the equation of the normal line

Using the point - slope form of a line $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(5,0)$ and $m=\frac{1}{25}$, we get $y-0=\frac{1}{25}(x - 5)$, which simplifies to $y=\frac{1}{25}x-\frac{1}{5}$.

Step5: Find the intersection points of the normal line and the parabola

Set the equation of the parabola equal to the equation of the normal line: $25x-5x^{2}=\frac{1}{25}x-\frac{1}{5}$. Multiply through by 25 to clear the fraction: $625x-125x^{2}=x - 5$. Rearrange to get a quadratic equation: $125x^{2}-624x - 5 = 0$. We know that $x = 5$ is one root (corresponding to the given point). We can use the fact that for a quadratic equation $ax^{2}+bx + c = 0$ ($a = 125$, $b=-624$, $c = - 5$), if the roots are $x_1$ and $x_2$, then $x_1+x_2=-\frac{b}{a}$. Since $x_1 = 5$, and $x_1+x_2=\frac{624}{125}$, then $5+x_2=\frac{624}{125}$, and $x_2=\frac{624}{125}-5=\frac{624 - 625}{125}=-\frac{1}{125}$.

Step6: Find the corresponding $y$ - value

Substitute $x=-\frac{1}{125}$ into the equation of the parabola $y = 25x-5x^{2}$. $y=25\times(-\frac{1}{125})-5\times(-\frac{1}{125})^{2}=-\frac{1}{5}-\frac{5}{15625}=-\frac{1}{5}-\frac{1}{3125}=-\frac{625 + 1}{3125}=-\frac{626}{3125}$.

Answer:

$(-\frac{1}{125},-\frac{626}{3125})$