# Find the equation of the parabola with vertex at origin, passing through the point P (3, - 4) and symmetric about the y-axis ?

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Find the equation of the parabola with vertex at origin, passing through the point P (3, - 4) and symmetric about the y-axis ?
1. 4x2 = -9y
2. 4x2 = 9y
3. x2 = -9y
4. None of these
5. x2 = - 3y

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Correct Answer - Option 1 : 4x2 = -9y

CONCEPT:

The following are the properties of a parabola of the form: x2 = - 4ay where a > 0

• Focus is given by (0, -a)
• Vertex is given by (0, 0)
• Equation of directrix is given by: y = a
• Equation of axis is given by: x = 0
• Length of latus rectum is given by: 4a
• Equation of latus rectum is given by: y = -a

CALCULATION:

Here, we have to find the equation of the parabola with vertex at origin, passing through the point (3, -4) and symmetric about the y-axis.

It is given that the vertex of the parabola is at (0, 0) and it is symmetric about the y-axis.

So, the equation of parabola can be either x2 = 4ay or x2 = -4ay

But since the parabola passes through the point P (3, - 4) and the point P lies in the 4th quadrant

i.e the parabola is facing downwards

So, the equation of the parabola is of the form x2 = -4ay

∵ The point P (3, -4) passes through the parabola so, x = 3 and y = -4 will satisfy the equation x2 = -4ay

⇒ 32 = - 4 ⋅ a ⋅ (-4)

⇒ a = 9/16

So, the required equation of parabola is: x2 = -4 ⋅ (9/16)y

⇒ 4x2 = -9y

Hence, option A is the correct answer.