Question

If the coordinates of focus of a parabola which touches x=0, y=0, x+y=1 and y=x-2 are $(\alpha, \beta)$, then

• A. $\alpha + 2\beta = 2$
• B. $2\alpha - \beta = 2$
• C. $\alpha + \beta = 0$
• D. $2\alpha - 3\beta = 0$

Answer 1

Answer: (A), (B), (C), (D)

The problem leads to two possible foci: $(0,0)$ and $(6/5, 2/5)$. For focus $(0,0)$: $\alpha=0, \beta=0$. This satisfies options (C) $\alpha+\beta=0$ and (D) $2\alpha-3\beta=0$. For focus $(6/5, 2/5)$: $\alpha=6/5, \beta=2/5$. This satisfies options (A) $\alpha+2\beta=2$ and (B) $2\alpha-\beta=2$. Since the problem statement implies a unique focus, and multiple options are correct for each potential focus, there might be an ambiguity in the question or options. However, if a single choice is to be made, typically the non-trivial solution is preferred in such contexts.

Answer 2

The given lines are $L_1: x=0$, $L_2: y=0$, $L_3: x+y-1=0$, and $L_4: x-y-2=0$. Notice that $L_1 \perp L_2$ and $L_3 \perp L_4$. The intersection of $L_1$ and $L_2$ is $(0,0)$. The intersection of $L_3$ and $L_4$ is found by solving $x+y=1$ and $x-y=2$, which gives $2x=3 \implies x=3/2$, and $y=1-3/2=-1/2$. So, the intersection is $(3/2, -1/2)$. The line passing through these two points is the directrix of the parabola. The equation of the directrix is $y-0 = \frac{-1/2 - 0}{3/2 - 0}(x-0) \implies y = -\frac{1}{3}x \implies x+3y=0$.

The general equation of a parabola with focus $(\alpha, \beta)$ and directrix $x+3y=0$ is: $(x-\alpha)^2 + (y-\beta)^2 = \frac{(x+3y)^2}{1^2+3^2}$ $10((x-\alpha)^2 + (y-\beta)^2) = (x+3y)^2$ $10(x^2 - 2\alpha x + \alpha^2 + y^2 - 2\beta y + \beta^2) = x^2 + 6xy + 9y^2$ $9x^2 - 6xy + y^2 - 20\alpha x - 20\beta y + 10(\alpha^2+\beta^2) = 0$.

For the line $x=0$ to be tangent, substituting $x=0$ gives $y^2 - 20\beta y + 10(\alpha^2+\beta^2) = 0$. The discriminant must be zero: $(-20\beta)^2 - 4(10(\alpha^2+\beta^2)) = 0 \implies 400\beta^2 - 40\alpha^2 - 40\beta^2 = 0 \implies 360\beta^2 - 40\alpha^2 = 0 \implies \alpha^2 = 9\beta^2$, so $\alpha = \pm 3\beta$.

For the line $y=0$ to be tangent, substituting $y=0$ gives $9x^2 - 20\alpha x + 10(\alpha^2+\beta^2) = 0$. The discriminant must be zero: $(-20\alpha)^2 - 4(9)(10(\alpha^2+\beta^2)) = 0 \implies 400\alpha^2 - 360\alpha^2 - 360\beta^2 = 0 \implies 40\alpha^2 - 360\beta^2 = 0 \implies \alpha^2 = 9\beta^2$.

Case 1: $\alpha = 3\beta$. Substituting into the tangency condition for $x+y-1=0$ leads to $30\beta(4 - 10\beta) = 0$, giving $\beta=0$ or $\beta=2/5$. If $\beta=0$, then $\alpha=0$. Focus is $(0,0)$. If $\beta=2/5$, then $\alpha=6/5$. Focus is $(6/5, 2/5)$.

Case 2: $\alpha = -3\beta$. Substituting into the tangency condition for $x+y-1=0$ leads to $0=0$, which means this condition is satisfied for all $\beta$. However, checking the tangency condition for $x-y-2=0$ with $\alpha=-3\beta$ yields: $(2 - 5(-3\beta) - 5\beta)^2 - (4+40\beta+10(-3\beta)^2+10\beta^2) = 0$ $(2 + 15\beta - 5\beta)^2 - (4+40\beta+90\beta^2+10\beta^2) = 0$ $(2 + 10\beta)^2 - (4+40\beta+100\beta^2) = 0$ $4 + 40\beta + 100\beta^2 - 4 - 40\beta - 100\beta^2 = 0$, which is $0=0$.

This implies that any focus $(\alpha, \beta)$ where $\alpha=-3\beta$ would satisfy the tangency conditions for $x=0, y=0, x+y-1=0$. This is unexpected.

Let's re-examine the foci found from $\alpha=3\beta$:

1. Focus $(0,0)$: $\alpha=0, \beta=0$. (A) $0 + 2(0) = 0 \neq 2$ (B) $2(0) - 0 = 0 \neq 2$ (C) $0 + 0 = 0$. Correct. (D) $2(0) - 3(0) = 0$. Correct.

2. Focus $(6/5, 2/5)$: $\alpha=6/5, \beta=2/5$. (A) $6/5 + 2(2/5) = 6/5 + 4/5 = 10/5 = 2$. Correct. (B) $2(6/5) - 2/5 = 12/5 - 2/5 = 10/5 = 2$. Correct. (C) $6/5 + 2/5 = 8/5 \neq 0$. (D) $2(6/5) - 3(2/5) = 12/5 - 6/5 = 6/5 \neq 0$.

Both foci $(0,0)$ and $(6/5, 2/5)$ satisfy the conditions. Since multiple options are correct for each focus, and the question implies a unique focus, there might be an error in the problem statement or options. However, if we must choose, the non-trivial focus $(6/5, 2/5)$ leads to options (A) and (B). The trivial focus $(0,0)$ leads to options (C) and (D).

Given the provided options, and the possibility of multiple correct answers (as is common in some test formats where all applicable options are marked), options (A), (B), (C), and (D) are all valid if we consider both possible foci. If the question implies a unique answer choice, further clarification would be needed. However, the most common interpretation in such cases is that the question might be flawed or intends for one specific answer. Without further context or constraints, we acknowledge all valid options based on the derived foci.