Question

what curve does $\sqrt{x - 6} + \sqrt{y - 5} = 4$ represent?

• A. Parabola
• B. Ellipse
• C. Hyperbola
• D. none

Answer 1

Answer: (A)

Parabola

Answer 2

Let $X = x - 6$ and $Y = y - 5$. The equation transforms to $\sqrt{X} + \sqrt{Y} = 4$. For the square roots to be defined, $X \ge 0$ and $Y \ge 0$. Rearrange the equation: $\sqrt{Y} = 4 - \sqrt{X}$. For $\sqrt{Y}$ to be real, $4 - \sqrt{X} \ge 0 \implies \sqrt{X} \le 4 \implies X \le 16$. Similarly, $Y \le 16$. Square both sides: $Y = (4 - \sqrt{X})^2 = 16 - 8\sqrt{X} + X$. Isolate the remaining square root term: $8\sqrt{X} = 16 + X - Y$. Square both sides again: $(8\sqrt{X})^2 = (16 + X - Y)^2$. $64X = 256 + 32(X - Y) + (X - Y)^2$. $64X = 256 + 32X - 32Y + X^2 - 2XY + Y^2$. Rearranging the terms, we get: $X^2 + Y^2 - 2XY - 32X - 32Y + 256 = 0$. This is a general second-degree equation of the form $AX^2 + BXY + CY^2 + DX + EY + F = 0$. Here, $A = 1$, $B = -2$, $C = 1$. The discriminant of the conic section is $B^2 - 4AC$. $B^2 - 4AC = (-2)^2 - 4(1)(1) = 4 - 4 = 0$. Since the discriminant is $0$, the curve represented is a Parabola.