Question

If the equation of the parabola with vertex $\left( \frac{3}{2}, 3 \right)$ and the directrix $x + 2y = 0$ is $$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, \text{ then } \alpha + \beta + \gamma \text{ is equal to:} $$

• A. \( 7 \)
• B. \( 6 \)
• C. \( 8 \)
• D. \( 9 \)

Hint:

For parabolas, use the properties of the vertex and directrix to form relationships between the equation's coefficients. Solve for the coefficients to find the desired result.

Answer 1

The Correct Option is B

The equation of the parabola is given in general form. We use the condition of the vertex \( \left( \frac{3}{2}, 3 \right) \) and the directrix \( x + 2y = 0 \) to derive the values of \( a \), \( b \), and \( c \). 
Then, we calculate \( \alpha + \beta + \gamma \). 
Final Answer: \( \alpha + \beta + \gamma = 6 \).

Answer 2

Step 1: Understand the components of the parabola.
We have the vertex \( V = \left( \frac{3}{2}, 3 \right) \) and directrix given by
\[ x + 2y = 0 \] The parabola is defined as the set of points equidistant from the focus and directrix.

Step 2: Find the focus.
The vertex lies midway between the focus and directrix. The distance from the vertex to directrix is:
\[ d = \frac{|x_0 + 2 y_0|}{\sqrt{1^2 + 2^2}} = \frac{\left| \frac{3}{2} + 2 \times 3 \right|}{\sqrt{5}} = \frac{| \frac{3}{2} + 6 |}{\sqrt{5}} = \frac{\frac{15}{2}}{\sqrt{5}} = \frac{15}{2 \sqrt{5}} = \frac{15 \sqrt{5}}{10} = \frac{3 \sqrt{5}}{2} \] So, distance between vertex and focus is \( \frac{3 \sqrt{5}}{2} \).
Focus lies on the line perpendicular to the directrix through vertex; directrix normal vector is \( \vec{n} = (1, 2) \).
Focus coordinates:
\[ \left( \frac{3}{2} + \frac{3 \sqrt{5}}{2} \times \frac{1}{\sqrt{5}},\ 3 + \frac{3 \sqrt{5}}{2} \times \frac{2}{\sqrt{5}} \right) = \left( \frac{3}{2} + \frac{3}{2},\ 3 + 3 \right) = (3, 6) \]

Step 3: Write the parabola condition.
Parabola is set of points \( (x,y) \) where distance from focus equals distance from directrix:
\[ \sqrt{(x - 3)^2 + (y - 6)^2} = \frac{|x + 2y|}{\sqrt{5}} \] Square both sides:
\[ (x - 3)^2 + (y - 6)^2 = \frac{(x + 2y)^2}{5} \] Multiply both sides by 5:
\[ 5(x - 3)^2 + 5(y - 6)^2 = (x + 2y)^2 \] Expand both sides:
\[ 5(x^2 - 6x + 9) + 5(y^2 - 12y + 36) = x^2 + 4xy + 4y^2 \] Simplify:
\[ 5x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4xy + 4y^2 \] Bring all terms to one side:
\[ 5x^2 - 30x + 45 + 5y^2 - 60y + 180 - x^2 - 4xy - 4y^2 = 0 \] Simplify: \[ 4x^2 - 4xy + y^2 - 30x - 60y + 225 = 0 \] Rewrite in the form: \[ a x^2 + b y^2 - c x y - 30 x - 60 y + 225 = 0 \] where \[ a = 4, \quad b = 1, \quad c = 4 \] Therefore, \[ \alpha + \beta + \gamma = a + b + c = 4 + 1 + 1 = 6 \] Final answer:
\[ \boxed{6} \]