Find the point on the graph of $y=x^2$ that is closest to the point $(0, 3)$
Here is Prob. 45, Sec. 2.3, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:
Find the point on the graph of $y = x^2$ that is closest to the point $(0, 3)$. ...
My Attempt:
Let $\left( a, a^2 \right)$ be the required point. Then the line through the points $(0, 3)$ and $\left( a, a^2 \right)$ is perpendicular to the line tangent to the curve $y = x^2$ at $\left( a, a^2 \right)$, and since the slope of the latter is $2a$, the slope of the former must satisfy$$ \frac{ a^2 - 3}{ a - 0} = -\frac{1}{2a}, $$provided of course that $a \neq 0$, which implies that$$ 2a^3 - 6a = -a, $$which implies that$$ 2a^3 = 5a, $$and as $a \neq 0$, so we must have$$ a^2 = 5/2, $$and hence $a = \pm \sqrt{5/2}$.
Let us compute the distance between the points $(0, 3)$ and $(\pm \sqrt{5/2}, 5/2 )$.$$ d \big( (0, 3), ( \pm \sqrt{5/2}, 5/2) \big) = \sqrt{ 5/2 + (5/2 - 3)^2 } = \sqrt{11}/2. $$
For $a = 0$, we note that the distance between the point $(0, 3)$ and the point $(0, 0)$ on the graph of $y = x^2$ is $3$.
Since $\sqrt{11}/2 < \sqrt{36}/2 = 6/2 = 3$, we can conclude that the point on the graph of the parabola $y = x^2$ that is closest to the point $(0, 3)$ is either of the two points $( \pm \sqrt{5/2}, 5/2 )$.
Is my solution correct in each and every detail? Or, have I made any mistakes?
$\endgroup$ 11 Answer
$\begingroup$its correct and rigorous but can be simplified ...
we have to minimise $$f(a)=\sqrt{a^2+{(a^2-3)}^2}$$ or basically minimise $$g(a)=a^2+{(a^2-3)}^2$$ $$g(a)={\left(a^2-\frac{5}{2}\right)}^2+\frac{11}{4}$$ which happens when $$a^2=\frac{5}{2} \iff a=\pm \sqrt{\frac{5}{2}}$$ and we are done!
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