Question:
If the equation of the parabola with vertex $\left(\frac{3}{2},\, 3\right)$ and the directrix $x + 2y = 0$ is
$$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, $$ then $\alpha + \beta + \gamma$ is equal to:
If the equation of the parabola with vertex $\left(\frac{3}{2},\, 3\right)$ and the directrix $x + 2y = 0$ is
$$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, $$ then $\alpha + \beta + \gamma$ is equal to:
$$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, $$ then $\alpha + \beta + \gamma$ is equal to:
Show 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.
Updated On: Mar 25, 2026
- \( 7 \)
- \( 6 \)
- \( 8 \)
- \( 9 \)
Show Solution
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The Correct Option is B
Solution and Explanation
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: \( \alpha + \beta + \gamma = 6 \).
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