Single-Variable-Calculus · Unit 13 · Video 3 · Interactive Practice

Slopes Into Shapes: The Parabolas and the Ellipses That Cross Them

IKey Formulas

Formula	Name	Video example
$\dfrac{dy}{dx} = \dfrac{2y}{x} \;\Rightarrow\; y = a x^2$	Tangent twice as steep as the ray	$\dfrac{dy}{y} = \dfrac{2\,dx}{x}$ , $\ln\lvert y\rvert = 2\ln\lvert x\rvert + c$ , $a$ any constant
$m_\perp = -\dfrac{1}{m}$	Perpendicular slopes	$-\dfrac{1}{2y/x} = -\dfrac{x}{2y}$
$\dfrac{dy}{dx} = -\dfrac{x}{2y} \;\Rightarrow\; \dfrac{x^2}{2} + y^2 = a^2$	Orthogonal trajectories: ellipses	semi-axes $\sqrt{2}\,a$ (horizontal) and $a$ (vertical)
$y = \pm\sqrt{a^2 - \tfrac{x^2}{2}}$	Explicit solutions: half-ellipses	each ends at $x = \pm\sqrt{2}\,a$ with a vertical tangent

Key Insight: Separation of variables can lose solutions (add back $y = 0$ ) or produce constants that give no curve (keep only $c > 0$ in $\tfrac{x^2}{2} + y^2 = c$ ).

IITangent Twice the Ray

Through every point off the $y$ -axis passes exactly one parabola $y = ax^2$ , and its tangent there is twice as steep as the ray.

IIIHalf-Parabolas Glued at the Origin

Because $2y/x$ is undefined on $x = 0$ , a left half of one parabola can join the right half of another.

Left half Right half x −1.00

💡 Knowing $y(-1) = 1$ fixes only the left half: the equation specifies no slope on $x = 0$ , so nothing ties the right half to it.

IVEllipses That Cross Every Parabola

At every point the ellipse slope $-x/(2y)$ is the negative reciprocal of the parabola slope $2y/x$ .

VQuiz Questions

Problem 1 · The Parabola Through a Point

Given: the general solution $y = ax^2$ of $\dfrac{dy}{dx} = \dfrac{2y}{x}$ — find the solution curve through $(2, 3)$ .

✅ Correct!

3 = a \cdot 2^2

gives

a = \tfrac34

; the tangent slope there is

2ax = 3

, twice the ray slope

\tfrac32

❌ Not quite. That divides

3

x = 2

; the family is

y = ax^2

, so divide by

x^2 = 4

❌ Not quite. That is

x^2/y

, upside down: solve

3 = 4a

for

a

❌ That is the ray. The line

y = \tfrac32 x

has slope

y/x

, not

2y/x

, so it fails the equation.

Show solution

Substitute $(x, y) = (2, 3)$ into $y = ax^2$ :

3 = a\cdot 2^2 = 4a \quad\Rightarrow\quad a = \tfrac{3}{4}, \qquad y = \tfrac{3}{4}x^2.

Check: $\dfrac{dy}{dx} = \tfrac{3}{2}x$ , which is $3$ at $x = 2$ , and $\dfrac{2y}{x} = \dfrac{2\cdot 3}{2} = 3$ . The ray from the origin to $(2, 3)$ has slope $\tfrac{3}{2}$ , half of $3$ .

Problem 2 · Which Constants Give Curves?

Given: separating $\dfrac{dy}{dx} = -\dfrac{x}{2y}$ gives $\dfrac{x^2}{2} + y^2 = c$ — find the values of $c$ for which this equation describes a curve.

✅ Correct! A sum of squares is never negative, and

c = 0

forces

x = y = 0

; writing

c = a^2

with

a > 0

gives the ellipses.

❌ Not quite. Unlike

y = Ax^2

, where

A

could have any sign, here the left side

\tfrac{x^2}{2} + y^2

is never negative.

❌ Close, but…

\tfrac{x^2}{2} + y^2 = 0

holds only at the single point

(0, 0)

, which is not a curve.

❌ Not quite. For

c < 0

no point satisfies the equation:

\tfrac{x^2}{2} + y^2 \geq 0

always.

Show solution

Both terms of $\tfrac{x^2}{2} + y^2$ are squares (times a positive number), so the left side is $\geq 0$ , and it equals $0$ only when $x = 0$ and $y = 0$ .

$c < 0$ : no points at all.
$c = 0$ : only the origin.
$c > 0$ : write $c = a^2$ ; then $\tfrac{x^2}{2} + y^2 = a^2$ is an ellipse.

So the method produced constants that must be taken away — the opposite of the earlier examples, where a lost solution ( $y = 0$ ) had to be added back.

Problem 3 · The Ellipse Through a Point

Given: the ellipses $\dfrac{x^2}{2} + y^2 = a^2$ — find the semi-axes of the one through $(2, 1)$ .

Horizontal semi-axis

Vertical semi-axis

✅ Correct!

a^2 = \tfrac{4}{2} + 1 = 3

, so the semi-axes are

\sqrt{2}\cdot\sqrt{3} = \sqrt{6}

and

\sqrt{3}

, in ratio

\sqrt{2}

❌ That uses $a^2$ instead of $a$ . The point gives

a^2 = 3

; the semi-axes are

\sqrt{2}\,a

and

a

with

a = \sqrt{3}

❌ Check the horizontal semi-axis. Set

y = 0

x^2 = 2a^2

, so

x = \sqrt{2}\,a

— the longer of the two.

❌ Check the vertical semi-axis. Set

x = 0

y^2 = a^2

, so

y = a

— the shorter of the two.

Show solution

Step 1: find $a^2$ by substituting $(2, 1)$ :

a^2 = \frac{2^2}{2} + 1^2 = 2 + 1 = 3, \qquad a = \sqrt{3}.

Step 2: horizontal semi-axis ( $y = 0$ ):

\frac{x^2}{2} = 3 \;\Rightarrow\; x^2 = 6 \;\Rightarrow\; x = \sqrt{6} = \sqrt{2}\,a.

Step 3: vertical semi-axis ( $x = 0$ ):

y^2 = 3 \;\Rightarrow\; y = \sqrt{3} = a.

The ratio is $\sqrt{6}/\sqrt{3} = \sqrt{2}$ , as for every ellipse in the family.

Problem 4 · Orthogonal Trajectories of Cubics

Given: the family $y = ax^3$ — find the family of curves that crosses every member at right angles.

✅ Correct! The cubics have slope

3y/x

, the new curves

-x/(3y)

, and separating

3y\,dy = -x\,dx

gives

x^2 + 3y^2 = C

❌ Not quite. Its slope is

-3x/y

; times

3y/x

that is

-9

, not

-1

. Recheck which variable gets the

3

❌ Close, but… That takes the reciprocal

x/(3y)

and drops the minus sign; these hyperbolas meet the cubics with slope product

+1

❌ Not quite. That negates the slope to

-3y/x

without taking the reciprocal; perpendicular slopes multiply to

-1

Show solution

Step 1: the family's differential equation. From $y = ax^3$ , $\dfrac{dy}{dx} = 3ax^2 = \dfrac{3ax^3}{x} = \dfrac{3y}{x}$ .

Step 2: perpendicular slope.

\frac{dy}{dx} = -\frac{1}{3y/x} = -\frac{x}{3y}.

Step 3: separate and integrate.

3y\,dy = -x\,dx \;\Rightarrow\; \tfrac{3}{2}y^2 = -\tfrac{1}{2}x^2 + c \;\Rightarrow\; x^2 + 3y^2 = C, \quad C = 2c > 0.

Check: implicit differentiation of $x^2 + 3y^2 = C$ gives $2x + 6y\,y' = 0$ , so $y' = -\dfrac{x}{3y}$ , and $\dfrac{3y}{x}\cdot\left(-\dfrac{x}{3y}\right) = -1$ . These are ellipses with horizontal semi-axis $\sqrt{3}$ times the vertical one.

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