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Solving the Assessment

From Formula Sheet to Excellence Proof

📜 NCEA Formula Sheet (Reference)

Ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ $$c^2 = a^2 - b^2 \text{ (if } a > b \text{)}$$ $$\text{Foci: } (\pm c, 0)$$
Hyperbola
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$c^2 = a^2 + b^2$$ $$\text{Foci: } (\pm c, 0)$$

*Note: The assessment defines $b^2 = a^2 - c^2$ for Ellipse and $b^2 = c^2 - a^2$ for Hyperbola depending on arrangement. Use the geometric relationship: $a$ is always the major axis (ellipse) or transverse axis (hyperbola).

Visual Check

Look at the critical points labeled on the graph below.

📝 The Assessment Task

"An ellipse with foci at (2, 0) and (-2, 0) passes through the point (4, 0)."

"A hyperbola also has foci at (2, 0) and (-2, 0) and passes through the point (1, 0)."

"Investigate the relationship between the gradient of the ellipse and the hyperbola at a point of intersection between the given ellipse and hyperbola."

Excellence Extension: "Investigate the relationship... where the foci are at $(c, 0)$ and $(-c, 0)$, the ellipse goes through $(2c, 0)$ and the hyperbola goes through $(c/2, 0)$."

🗺️ Strategy Map: How to break it down

We will decompose this big paragraph into 3 clear tasks:

  • Task 1 (Achieved): Use the formula sheet to build the equations for the specific case.
  • Task 2 (Merit): Find where they cross, calculate gradients ($dy/dx$), and see if they are perpendicular.
  • Task 3 (Excellence): Do it all again, but using algebra ($c$) instead of numbers.

Task 1: Build the Equations

Ellipse: Passes through $(4,0) \rightarrow a=4$. Foci at $c=2$.

Use $b^2 = a^2 - c^2$:

Hyperbola: Passes through $(1,0) \rightarrow a=1$. Foci at $c=2$.

Use $b^2 = c^2 - a^2$:

Task 2: The Investigation

Goal: Find the angle between the curves.

Equations Found:

  • Ellipse: $3x^2 + 4y^2 = 48$
  • Hyperbola: $3x^2 - y^2 = 3$

Step 1: Find Intersections

Solve the simultaneous equations. (Hint: Multiply the Hyperbola equation by 4 to eliminate y).

Step 2: Implicit Differentiation

We need $dy/dx$ for both curves.

Ellipse ($3x^2 + 4y^2 = 48$): $6x + 8y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-3x}{4y}$

Hyperbola ($3x^2 - y^2 = 3$): $6x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{3x}{y}$

Step 3: The Product of Gradients

At intersection $(2, 3)$, multiply the two gradients.

Task 3: The General Case

Goal: Prove they are ALWAYS perpendicular.

Parameters: Foci $(\pm c, 0)$. Ellipse goes through $2c$. Hyperbola goes through $c/2$.

Step 1: Define Parameters in terms of $c$

Ellipse: $a = 2c$. Find $b^2$.

$b^2 = a^2 - c^2 = (2c)^2 - c^2 = 3c^2$.

Equation: $\frac{x^2}{4c^2} + \frac{y^2}{3c^2} = 1 \Rightarrow 3x^2 + 4y^2 = 12c^2$.

Hyperbola: $a = c/2$. Find $b^2$.

$b^2 = c^2 - a^2 = c^2 - (c/2)^2 = \frac{3c^2}{4}$.

Equation: $\frac{x^2}{c^2/4} - \frac{y^2}{3c^2/4} = 1 \Rightarrow 12x^2 - 4y^2 = 3c^2$.

Step 2: Interactive Proof

Let's verify the gradients again with these general forms.

Ellipse Gradient: $m_E = \frac{-3x}{4y}$

Hyperbola Gradient: $m_H = \frac{12x}{4y} = \frac{3x}{y}$

Multiply them: $m_E \times m_H = \left(\frac{-3x}{4y}\right) \left(\frac{3x}{y}\right) = \frac{-9x^2}{4y^2}$

The Final Logic: At intersection, we found $y = \frac{3}{2}x$. Substitute this into the product:

$$ \text{Product} = \frac{-9x^2}{4(\frac{3}{2}x)^2} = \frac{-9x^2}{4(\frac{9}{4}x^2)} = \frac{-9x^2}{9x^2} = -1 $$

Since the product is -1, the curves are Orthogonal (Perpendicular) at every intersection!

Explore the General Case

Is it a Circle? No, it just looks round!
Ellipse Width = $2c$. Height = $\sqrt{3}c \approx 1.73c$.
Look at the labeled intercepts below to see the difference.

*Note: The aspect ratio is locked to ensure the perpendicular lines look correct (90°).