Chapter 6 · Application of Derivatives

Exercise 6.3 · Q27 (MCQ): Point on $x^2 = 2y$ nearest to $(0, 5)$

Exercise 6.3Optimisation · Section 6.4

The question

Choose the correct answer in Questions 27 and 29.

27. The point on the curve $x^2 = 2y$ which is nearest to the point $(0, 5)$ is

(A) $\left(2\sqrt{2}, 4\right)$
(B) $\left(2\sqrt{2}, 0\right)$
(C) $(0, 0)$
(D) $(2, 2)$

Exam answer

A point of $x^2 = 2y$ is $\left(x, \frac{x^2}{2}\right)$ ⁠. Its squared distance from $(0, 5)$ is

\begin{aligned} D &= x^2 + \left(\frac{x^2}{2} - 5\right)^2 \\ &= \frac{x^4}{4} - 4x^2 + 25 \end{aligned}

$D'(x) = x^3 - 8x = 0$ gives $x^2 = 8$ ⁠, where $D''(x) = 3x^2 - 8 = 16 > 0$ ⁠. So $D$ is least, $D = 9$ ⁠, at $x^2 = 8$ ⁠, $y = 4$ ⁠.

Answer: (A) $\left(2\sqrt2, 4\right)$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Every point of the curve can be written using one number, its $x$ -coordinate: $\left(x, \frac{x^2}{2}\right)$ ⁠. The distance from $(0, 5)$ is then a function of $x$ ⁠, and the distance is least exactly where its square is least, which avoids a square root. The answer is not the point of the parabola straight below $(0, 5)$ ⁠: the vertex $(0, 0)$ is in fact a local maximum of the distance.

Given, to find, and the plan

Given: The parabola $x^2 = 2y$ and the point P $(0, 5)$ ⁠.
To find: The option that is the point of the parabola nearest to P.
Plan: Write a general point of the curve in terms of $x$ ⁠; write the squared distance $D(x)$ ⁠; find its critical points and classify them; prove the least value for all real $x$ ⁠; then examine every option.

Basics you need

1Distance formula

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ ⁠.

2Minimise the square of a distance

Distances are never negative, and for $a, b \ge 0$ ⁠, $a < b$ exactly when $a^2 < b^2$ ⁠. So the nearest point is the one with the least squared distance. Basics: building the function to optimise

3Factorising

$x^3 - 8x = x(x^2 - 8)$ $= x\left(x - 2\sqrt{2}\right)\left(x + 2\sqrt{2}\right)$ ⁠, since $8 = \left(2\sqrt{2}\right)^2$ ⁠. Basics: factorising and the discriminant

4Second derivative test (Theorem 4, Section 6.4)

$D'(c) = 0$ with $D''(c) > 0$ ⁠: local minimum; with $D''(c) < 0$ ⁠: local maximum. Lesson: first and second derivative tests

Solution

A general point of the curve

From $x^2 = 2y$ we get $y = \frac{x^2}{2}$ ⁠. So the points of the curve are exactly
$\left(x, \tfrac{x^2}{2}\right), \qquad x \in \mathbf{R}.$

The parabola $x^2 = 2y$ and P $(0, 5)$ ⁠, equal scales on both axes. The circle of radius 3 about P just touches the parabola at $\left(\pm 2\sqrt{2}, 4\right)$ ⁠; option A is the right-hand point. B is not on the curve; C and D are on it but outside the circle, so farther from P.

The curve is a single family of points described by one free number $x$ ⁠, so the distance to P becomes a function of one variable. Every real $x$ is allowed.

Solving the curve's equation for $y$ is possible here because $y$ appears only once and to the first power.
The squared distance as a function of $x$
$\begin{aligned} D(x) &= (x - 0)^2 + \left(\tfrac{x^2}{2} - 5\right)^2 \\ &= x^2 + \tfrac{x^4}{4} - 5x^2 + 25 \\ &= \tfrac{x^4}{4} - 4x^2 + 25. \end{aligned}$
The distance itself is $\sqrt{D(x)}$ ⁠, and it is least exactly where $D(x)$ is least (Basic 2).

Working with $D$ instead of $\sqrt{D}$ gives a polynomial, which is much easier to differentiate, and does not change where the minimum is.

$\left(\frac{x^2}{2} - 5\right)^2 = \frac{x^4}{4} - 5x^2 + 25$ ⁠. Basics: building the function to optimise
Critical points
$D'(x) = x^3 - 8x = x\left(x - 2\sqrt{2}\right)\left(x + 2\sqrt{2}\right).$
So $D'(x) = 0$ at $x = 0$ ⁠, $x = 2\sqrt{2}$ and $x = -2\sqrt{2}$ ⁠. $D$ is a polynomial, so there are no other critical points.

The least value of $D$ on the whole real line, if it exists, must be at a critical point (Theorem 2, Section 6.4), so these three are the only candidates.

Power rule; factor out $x$ ⁠, then use $a^2 - b^2 = (a - b)(a + b)$ (Basic 3). Basics: factorising

Classify them with the second derivative

D''(x) = 3x^2 - 8.

$x$	$D''(x)$	Type	$D(x)$
$0$	$-8 < 0$	local maximum	25
$2\sqrt{2}$	$24 - 8 = 16 > 0$	local minimum	$16 - 32 + 25 = 9$
$-2\sqrt{2}$	$16 > 0$	local minimum	9

The vertex

(0, 0)

⁠, straight below P, is a local maximum of the distance. Moving sideways a small amount

\delta

along the parabola adds about

\delta^2

D

⁠, but the curve also rises by

\frac{\delta^2}{2}

towards P, which removes about

5\delta^2

⁠: overall

D \approx 25 - 4\delta^2

⁠, smaller than 25.

Theorem 4 (i) and (ii), Section 6.4. At

x^2 = 8

⁠:

\frac{x^4}{4} = \frac{64}{4} = 16

and

4x^2 = 32

⁠. Lesson: first and second derivative tests

The least distance for all real $x$

Complete the square in $x^2$ ⁠:
$D(x) = \frac{(x^2 - 8)^2}{4} + 9 \ \ge\ 9,$
with equality exactly when $x^2 = 8$ ⁠, that is $x = \pm 2\sqrt{2}$ ⁠. (Check: $\frac{(x^2 - 8)^2}{4} = \frac{x^4}{4} - 4x^2 + 16$ ⁠, and adding 9 gives $D(x)$ ⁠.) So the least distance is $\sqrt{9} = 3$ ⁠, reached at the points
$\left(2\sqrt{2}, 4\right) \quad\text{and}\quad \left(-2\sqrt{2}, 4\right),$
since $y = \frac{x^2}{2} = \frac{8}{2} = 4$ ⁠.

Local minima only compare nearby points; the whole curve is infinite, so we need an argument for every $x$ ⁠. Writing $D$ as a square plus 9 does that in one line: a square is never negative.

A square of a real number is $\ge 0$ ⁠, and equals 0 only when the number is 0. Lesson: optimisation problems

Examine every option

A candidate must first lie on the curve ( $x^2 = 2y$ ⁠); then its distance from P is compared with the least distance 3.

Option	On $x^2 = 2y$ ⁠?	Distance to $(0, 5)$	Verdict
(A) $\left(2\sqrt{2}, 4\right)$	$8 = 8$ ⁠: yes	$\sqrt{8 + 1} = 3$	nearest: correct
(B) $\left(2\sqrt{2}, 0\right)$	$8 \ne 0$ ⁠: no	$\sqrt{8 + 25} = \sqrt{33} \approx 5.7446$	not a point of the curve
(C) $(0, 0)$	$0 = 0$ ⁠: yes	$5$	local maximum of the distance (Step 4)
(D) $(2, 2)$	$4 = 4$ ⁠: yes	$\sqrt{4 + 9} = \sqrt{13} \approx 3.6056$	on the curve but farther than 3

Each wrong option fails for a different reason. (B) has the right

x

-coordinate but is not on the curve at all. (C) is a critical point, but of the wrong kind. (D) is on the curve, between C and A, but

\sqrt{13} > 3

⁠.

Distance formula (Basic 1) with P

(0, 5)

⁠.

Result

The nearest points of $x^2 = 2y$ to $(0, 5)$ are $\left(\pm 2\sqrt{2}, 4\right)$ ⁠, at distance 3. Of the options, only $\left(2\sqrt{2}, 4\right)$ is one of them: the answer is (A).

(Textbook answer: A. Agrees. The point $\left(-2\sqrt{2}, 4\right)$ is equally near but is not among the options.)

Check it with numbers

$D(x) = \frac{x^4}{4} - 4x^2 + 25$ at a few points: $D(2) = 4 - 16 + 25 = 13$ ⁠, $D(2.8284) \approx 9.0000$ ⁠, $D(3) = 20.25 - 36 + 25 = 9.25$ ⁠, $D(0) = 25$ ⁠. The smallest is at $x \approx 2.8284 = 2\sqrt{2}$ ⁠, where the distance is $\sqrt{9} = 3$ ⁠.

Common mistakes

Picking the vertex because it is "straight below" P. For $x^2 = 2y$ the vertex is nearest only when P $(0, c)$ has $c \le 1$ ⁠; for P $(0, 5)$ it is a local maximum of the distance.
Not checking that an option is on the curve. (B) has the same $x$ -coordinate as A, but it does not lie on $x^2 = 2y$ at all.
Stopping at the local minima. A local minimum of a function on the whole real line need not be the least value; the identity $D(x) = \frac{(x^2 - 8)^2}{4} + 9$ proves it is.
Differentiating the square root. It works, but the squared distance gives the same points with much less algebra.

What to take away

For "nearest point on a curve", minimise the squared distance from a general point of the curve.
Check every candidate in an MCQ against both conditions: on the curve, and at the least distance.
Example 23 in Section 6.4 treats the point $(0, c)$ and the parabola $y = x^2$ ⁠; this question is the same idea with a different parabola.

Every fact used, and where it comes from

Fact	Used in	Source
Distance formula	Steps 2, 6	Coordinate geometry (Class XI)
Least distance where the squared distance is least	Step 2	Algebra of inequalities (Basics: building the function)
Extrema only at critical points	Step 3	Theorem 2, Section 6.4
Second Derivative Test	Step 4	Theorem 4, Section 6.4 (Lesson: tests)
Completing the square	Step 5	Algebra (Basics: factorising)
Same method for $(0, c)$ and $y = x^2$	Takeaways	Example 23, Section 6.4

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Exercise 6.3 · Q27 (MCQ): Point on x2=2yx^2 = 2yx2=2y nearest to (0,5)(0, 5)(0,5)

Basics you need

Solution

A general point of the curve

The squared distance as a function of xxx

Critical points

Classify them with the second derivative

The least distance for all real xxx

Examine every option

Every fact used, and where it comes from

Exercise 6.3 · Q27 (MCQ): Point on $x^2 = 2y$ nearest to $(0, 5)$

The squared distance as a function of $x$

The least distance for all real $x$