Classifying a Conic From Its General Equation
An engineering student is given the equation 4x^2 - 9y^2 + 16x + 54y - 101 = 0 and told it models the cross-section of a cooling tower's support. She must identify the conic, find its centre and key features, and decide whether a circular reinforcing ring of radius 5 centred at the conic's centre would intersect the curve. A teammate claims the equation is a circle because both squared terms appear, and another claims it is an ellipse because the coefficients have opposite signs after moving terms. The student must settle the classification rigorously and answer the intersection question.
1. Complete the square to write the equation in standard form, identify the conic with its centre, vertices, and asymptotes, and explain why the teammates' claims are wrong.
Read the worked answer
Group terms: (4x^2 + 16x) + (-9y^2 + 54y) = 101.
Complete the square: 4(x^2 + 4x) = 4[(x + 2)^2 - 4] = 4(x + 2)^2 - 16. -9(y^2 - 6y) = -9[(y - 3)^2 - 9] = -9(y - 3)^2 + 81.
Substitute: 4(x + 2)^2 - 16 - 9(y - 3)^2 + 81 = 101, so 4(x + 2)^2 - 9(y - 3)^2 = 36.
Divide by 36: (x + 2)^2/9 - (y - 3)^2/4 = 1.
This is a hyperbola opening left and right, centred at (-2, 3), with a^2 = 9 (a = 3) and b^2 = 4 (b = 2). Vertices are at (-2 ± 3, 3): (1, 3) and (-5, 3). The asymptotes pass through the centre with slopes ±b/a = ±2/3: y - 3 = ±(2/3)(x + 2).
Why the teammates are wrong: a circle requires equal coefficients on x^2 and y^2 with no xy term; here the coefficients are 4 and -9, unequal. An ellipse requires both squared terms to have the same sign after completing the square; here the coefficients 4 and -9 have opposite signs, giving a difference equal to a positive constant, which is a hyperbola. Opposite signs in the original equation do not indicate an ellipse; they indicate a hyperbola.
2. Determine whether the circle (x + 2)^2 + (y - 3)^2 = 25 intersects the hyperbola, using an algebraic substitution rather than a sketch.
Read the worked answer
The circle is centred at (-2, 3) with radius 5, the same centre as the hyperbola. Substitute u = x + 2 and v = y - 3 to simplify: the hyperbola is u^2/9 - v^2/4 = 1 and the circle is u^2 + v^2 = 25.
From the circle, v^2 = 25 - u^2. Substitute into the hyperbola: u^2/9 - (25 - u^2)/4 = 1. Multiply by 36: 4u^2 - 9(25 - u^2) = 36, so 4u^2 - 225 + 9u^2 = 36, giving 13u^2 = 261, so u^2 = 261/13 ≈ 20.08, and u ≈ ±4.48.
Then v^2 = 25 - 20.08 = 4.92, so v ≈ ±2.22.
Since u^2 ≈ 20.08 > 9, the points lie beyond the vertices (which are at u = ±3), so they are on the hyperbola's branches. There are four intersection points: (u, v) ≈ (±4.48, ±2.22), i.e. (x, y) ≈ (2.48, 5.22), (2.48, 0.78), (-6.48, 5.22), (-6.48, 0.78).
Check: for u^2 = 261/13, u^2/9 = 261/117 ≈ 2.231 and v^2/4 = 4.92/4 ≈ 1.231; difference ≈ 1.000, correct. So the ring does intersect the curve at four points, and the algebraic method confirms it without relying on a sketch.