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The directrix of the parabola y=1/4x^2-2 is y =l. what is the value of I?

User RDL
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Final answer:

The value of the constant l for the directrix of the given parabola y = 1/4x² - 2 is -2.25.

Step-by-step explanation:

The question is asking for the value of the constant l in the equation for the directrix of a parabola. The given parabola is in the form y = ax² + bx + c, which for this case is y = ⅛ x² - 2. To find the directrix of this parabola, we use the fact that for a parabola y = ax² + bx + c, the directrix is y = c - ⅝ / (4a) when the parabola opens upwards. In our equation, a = ⅛ and c = -2.

Following the formula for the directrix:

Directrix, y = -2 - (⅝ / (⅛ × 4))
= -2 - (1/4)
= -2 - 0.25
= -2.25

So, the value of the constant l for the directrix y = l is -2.25.

User Blacktrance
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