As you can see I found the equation but I don't know how to find the points. As far as I tried was $(7, 49)$ but it was wrong.
[Math] Find the point on a parabola that is closest to a given point
calculusconic sectionsoptimization
Related Solutions
If you have a curve $y=f(x)$, the length of the curve between points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is given by $$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{(dx)^2 + (df(x))^2} = \int_{x_1}^{x_2}\sqrt{1+\left(\dfrac{df}{dx} \right)^2} dx$$ In your case, $f(x) = x^2$. Hence, $\dfrac{df}{dx} = 2x$. Hence, the length of curve between the points $(x_1,x_1^2)$ and $(x_2,x_2^2)$ is $$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{1+(2x)^2} dx = \int_{x_1}^{x_2} \sqrt{1+4x^2} dx$$ Can you evaluate this integral?
$$\int_{x_1}^{x_2} \sqrt{1+4x^2} dx = \left.\dfrac{2x \sqrt{4x^2+1} + \log(2x+\sqrt{1+4x^2})}4 \right \vert_{x_1}^{x_2}$$
EDIT
Added picture to highlight the different lengths.
Let $(x,y) = \left(\dfrac{y^2}{2},y\right)$ be the point on the parabola that is closest to $(1,0)$, then this point is where the distance of the two point is minimized. $d = \sqrt{\left(\dfrac{y^2}{2} - 1\right)^2 + y^2} = \sqrt{\dfrac{y^4}{4} + 1} \geq 1$, and is minimized when $y = 0$. Thus the sought point is the origin $(0,0)$.
Best Answer
Minimizing the function is the same as minimizing its square. You wrote the equation for the distance, but the exercise asks its square, so we would have $$s(x) = (x - 7)^2 + (y - 0)^2$$ But the point $(x,y)$ is in the parabola $y = x^2$. So we get $$s(x) = (x - 7)^2 + x^4$$
Now, can you solve $s '(x) = 0 $? This will give you the critical points of $s$.