4.3 Conic Sections: Circles & Parabolas in the Coordinate Plane

4.3 Conic Sections: Circles & Parabolas in the Coordinate Plane

Conic sections are planar curves formed by the intersection of a plane with a double right circular cone. On the Bocconi Admission Test, conic section questions focus primarily on circles and parabolas. Candidates must be proficient at completing the square to extract geometric invariants (center, radius, vertex, axis of symmetry) and applying tangency criteria between straight lines and quadratic curves.

---

1. The Circle in the Coordinate Plane

A circle is the locus of all points $(x, y) \in \mathbb{R}^2$ equidistant from a fixed center $C(h, k)$ by a positive radius $r > 0$.

Canonical Equation

Applying the distance formula $d(P, C) = r$ and squaring both sides gives the canonical (standard) equation:

$$(x - h)^2 + (y - k)^2 = r^2$$

Expanded General Form

Expanding the binomials:

$$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2 \implies x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$$

Setting $a = -2h$, $b = -2k$, and $c = h^2 + k^2 - r^2$, we arrive at the expanded general form:

$$x^2 + y^2 + ax + by + c = 0$$

Completing the Square Algorithm

Given an equation in expanded form, extract center and radius using standard relations:

1. Center Coordinates: $$h = -\frac{a}{2}, \qquad k = -\frac{b}{2} \implies C\left(-\frac{a}{2}, -\frac{b}{2}\right)$$
2. Radius: $$r^2 = h^2 + k^2 - c = \left(-\frac{a}{2}\right)^2 + \left(-\frac{b}{2}\right)^2 - c = \frac{a^2 + b^2 - 4c}{4}$$ $$r = \sqrt{\frac{a^2 + b^2 - 4c}{4}} = \frac{1}{2} \sqrt{a^2 + b^2 - 4c}$$

Existence Conditions for Real Circles

• If $a^2 + b^2 - 4c > 0$: The equation defines a real circle with non-zero radius $r > 0$.
• If $a^2 + b^2 - 4c = 0$: The equation defines a degenerate point circle consisting of the single point $(h, k)$.
• If $a^2 + b^2 - 4c < 0$: The equation has no real locus (empty set in $\mathbb{R}^2$).

Bocconi Exam Trap: If an equation has non-unit leading coefficients, such as $3x^2 + 3y^2 - 12x + 18y - 15 = 0$, you must divide the entire equation by 3 before reading coefficients $a, b, c$. Failing to normalize leading coefficients leads to incorrect centers and radii.

---

2. Lines and Circles: Intersections and Tangency Conditions

Determining whether a line $r: Ax + By + C = 0$ intersects a circle $(x - h)^2 + (y - k)^2 = r^2$ can be done geometrically or algebraically.

The Geometric Distance Method (Bocconi Fast Method)

Calculate the perpendicular distance $d$ from the circle's center $C(h, k)$ to the line:

$$d = \frac{|A h + B k + C|}{\sqrt{A^2 + B^2}}$$

• Secant Line ($d < r$): The line intersects the circle at two distinct points. The chord length is $2\sqrt{r^2 - d^2}$.
• Tangent Line ($d = r$): The line intersects the circle at exactly one point (point of tangency).
• Exterior Line ($d > r$): The line does not intersect the circle.

The Algebraic Discriminant Method

Substitute the line equation $y = mx + q$ into the circle equation, obtaining a single quadratic equation in $x$: $P(x) = 0$. The discriminant $\Delta$ governs the intersection:

• $\Delta > 0 \implies$ Secant (two real roots).
• $\Delta = 0 \implies$ Tangent (one repeated real root).
• $\Delta < 0 \implies$ Exterior (no real roots).

Tangent Line at a Point on the Circle

If point $P(x_0, y_0)$ lies on the circle $(x - h)^2 + (y - k)^2 = r^2$, the tangent line at $P$ is perpendicular to radius $CP$. Its equation is given by the splitting rule:

$$(x_0 - h)(x - h) + (y_0 - k)(y - k) = r^2$$

For a circle centered at the origin ($x^2 + y^2 = r^2$), this simplifies to:

$$x_0 x + y_0 y = r^2$$

---

3. Parabolas in the Coordinate Plane

A parabola is the geometric locus of all points in $\mathbb{R}^2$ equidistant from a fixed point (the focus $F$) and a fixed line (the directrix $d$).

Vertical Parabolas: $y = ax^2 + bx + c$ ($a \neq 0$)

Vertical parabolas have a vertical axis of symmetry and represent quadratic functions of $x$:

• Orientation:
  • If $a > 0$: The parabola opens upward (concave up). The vertex is the global minimum.
  • If $a < 0$: The parabola opens downward (concave down). The vertex is the global maximum.
• Vertex $V(x_V, y_V)$: $$x_V = -\frac{b}{2a}, \qquad y_V = -\frac{\Delta}{4a} = c - \frac{b^2}{4a} \quad [\text{where } \Delta = b^2 - 4ac]$$ Alternative: Once $x_V$ is found, evaluate $y_V = f(x_V)$ directly to avoid memorizing $-\frac{\Delta}{4a}$.
• Axis of Symmetry: The vertical line passing through the vertex: $$x = -\frac{b}{2a}$$
• Roots ($x$-Intercepts): Given by the quadratic formula $x = \frac{-b \pm \sqrt{\Delta}}{2a}$:
  • $\Delta > 0$: Two distinct $x$-intercepts (parabola crosses the $x$-axis).
  • $\Delta = 0$: Exactly one $x$-intercept at the vertex (parabola is tangent to the $x$-axis).
  • $\Delta < 0$: No real $x$-intercepts (parabola lies entirely above or below the $x$-axis).
• $y$-Intercept: Point $(0, c)$.
• Focus and Directrix:
  • Focus: $F\left(-\frac{b}{2a}, \frac{1 - \Delta}{4a}\right)$
  • Directrix: $y = -\frac{1 + \Delta}{4a}$

Vertex Form of a Parabola

Completing the square on $y = ax^2 + bx + c$ yields vertex form:

$$y = a(x - h)^2 + k$$

Here, the vertex is immediately identified as $V(h, k)$.

Horizontal Parabolas: $x = ay^2 + by + c$ ($a \neq 0$)

Horizontal parabolas open sideways and are not functions of $x$:

• If $a > 0$: Opens to the right.
• If $a < 0$: Opens to the left.
• Vertex: $V\left(-\frac{\Delta}{4a}, -\frac{b}{2a}\right)$.
• Axis of symmetry: Horizontal line $y = -\frac{b}{2a}$.

---

4. Intersections of Lines and Parabolas

To find the intersection between a line $y = mx + q$ and vertical parabola $y = ax^2 + bx + c$, equate the two expressions:

$$ax^2 + bx + c = mx + q \iff ax^2 + (b - m)x + (c - q) = 0$$

The nature of the intersection is determined by the discriminant of this resolving quadratic:

$$\Delta_{\text{system}} = (b - m)^2 - 4a(c - q)$$

• Secant Line ($\Delta_{\text{system}} > 0$): Intersects the parabola at two distinct points.
• Tangent Line ($\Delta_{\text{system}} = 0$): Touches the parabola at exactly one point.
• Exterior Line ($\Delta_{\text{system}} < 0$): No real intersection.

Calculus Shortcut for Tangents: The slope of the tangent line to the parabola $y = ax^2 + bx + c$ at any point $x_0$ is given by the derivative: $m = y'(x_0) = 2ax_0 + b$. This calculates the tangent slope in seconds without expanding $\Delta = 0$.

---

5. Worked Step-by-Step Examples

Example 1: Finding Circle Center, Radius, and Line Secancy

Consider the circle $x^2 + y^2 - 4x + 6y - 12 = 0$ and the line $3x + 4y - 19 = 0$. Determine if the line is secant, tangent, or exterior.

1. Find center and radius of the circle:
   • $a = -4, b = 6, c = -12$.
   • Center: $h = -(-4)/2 = 2$, $k = -(6)/2 = -3 \implies C(2, -3)$.
   • Radius: $r = \sqrt{2^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.
2. Compute perpendicular distance from center $C(2, -3)$ to line $3x + 4y - 19 = 0$: $$d = \frac{|3(2) + 4(-3) - 19|}{\sqrt{3^2 + 4^2}} = \frac{|6 - 12 - 19|}{\sqrt{25}} = \frac{|-25|}{5} = \frac{25}{5} = 5$$
3. Compare $d$ and $r$: Since $d = r = 5$, the line is strictly tangent to the circle.

Example 2: Parabola Reconstruction from Geometric Features

Find the equation of the parabola with vertical axis of symmetry having vertex $V(3, -2)$ and passing through point $P(5, 6)$.

1. Use vertex form: $y = a(x - h)^2 + k = a(x - 3)^2 - 2$.
2. Substitute coordinates of point $P(5, 6)$ to determine $a$: $$6 = a(5 - 3)^2 - 2 \implies 6 = a(2)^2 - 2 \implies 6 = 4a - 2 \implies 4a = 8 \implies a = 2$$
3. Expand to standard polynomial form: $$y = 2(x - 3)^2 - 2 = 2(x^2 - 6x + 9) - 2 = 2x^2 - 12x + 18 - 2 = 2x^2 - 12x + 16$$
4. Factored form: $y = 2(x - 2)(x - 4)$, showing roots at $x = 2$ and $x = 4$.

Example 3: Tangent Line to a Parabola with Specified Slope

Find the value of $q$ such that the line $y = 4x + q$ is tangent to the parabola $y = x^2 - 2x + 5$.

1. Equate line and parabola equations: $$x^2 - 2x + 5 = 4x + q \implies x^2 - 6x + (5 - q) = 0$$
2. Impose tangency condition $\Delta = 0$: $$\Delta = (-6)^2 - 4(1)(5 - q) = 36 - 20 + 4q = 16 + 4q = 0 \implies 4q = -16 \implies q = -4$$
3. Find point of tangency: With $q = -4$, $x^2 - 6x + 9 = 0 \implies (x - 3)^2 = 0 \implies x = 3$. Then $y = 4(3) - 4 = 8$. Point of tangency is $(3, 8)$.