Question

Vertex of

$x^2+4y^2-4xy-2x+1=0$

• A. The vertex is at (13/25, 4/25)
• B. The vertex is at (1/5, 0)
• C. The vertex is at (0, 1/2)
• D. The vertex is at (1, 0)

Answer 1

Answer: (A)

(13/25, 4/25)

Answer 2

The given equation is $x^2+4y^2-4xy-2x+1=0$. This can be rewritten as $(x - 2y)^2 - 2x + 1 = 0$. This is the equation of a parabola. The axis of the parabola is parallel to $x - 2y = 0$. The slope of the axis is $1/2$. The tangent at the vertex is perpendicular to the axis, so its slope is $-2$. Differentiating the equation implicitly with respect to $x$: $2x - 4(y + x\frac{dy}{dx}) + 8y\frac{dy}{dx} - 2 = 0$ $(8y - 4x)\frac{dy}{dx} = 2 - 2x + 4y$ $\frac{dy}{dx} = \frac{2 - 2x + 4y}{8y - 4x} = \frac{1 - x + 2y}{4y - 2x}$ Setting $\frac{dy}{dx} = -2$: $\frac{1 - x + 2y}{4y - 2x} = -2$ $1 - x + 2y = -8y + 4x$ $1 + 10y = 5x \implies x = 2y + \frac{1}{5}$ Substitute $x$ into the simplified equation: $((2y + \frac{1}{5}) - 2y)^2 - 2(2y + \frac{1}{5}) + 1 = 0$ $(\frac{1}{5})^2 - 4y - \frac{2}{5} + 1 = 0$ $\frac{1}{25} - 4y - \frac{10}{25} + \frac{25}{25} = 0$ $\frac{16}{25} = 4y \implies y = \frac{4}{25}$ Substitute $y$ back into the expression for $x$: $x = 2(\frac{4}{25}) + \frac{1}{5} = \frac{8}{25} + \frac{5}{25} = \frac{13}{25}$ The vertex is at $(\frac{13}{25}, \frac{4}{25})$.