Find the vertex, focus , and directrix axis of the parabola `x^2+4y+3x=2`. Sketch the curve.

Explanation

Given equation of parabola is
`2y^(2) + 3y - 4x -3 =0`
or `2(y^(2) + (3)/(2)y) = 4x +3`
or `2(y^(2)+(3)/(2)y+((3)/(2))^(2)-((3)/(4))^(2)) =4x+3`.
or `2(y+(3)/(4))^(2) = 4x + 3 + (9)/(8) = 4x + (33)/(8)`
or `(y + (3)/(4))^(2) = 2(x+(33)/(32))`
This is of the form `Y^(2) = 4ax`, where `4a = 2`
Hence equation of axis is `y + (3)/(4) = 0` or y = `-(3)/(4)`
Equation of tangent at the vertex is
`x + (33)/(32) = 0`
`rArr x = -(33)/(32)`
`therefore` vetex is `((-33)/(32),(-3)/(4))`
( To get the vetex solve the equation of axis and the tangent at the vertex)
Length of latus rectum ` = |4a| = |2| = 2`
Focus is given by X = a, Y = 0
i.e., `x+(33)/(32) = (1)/(2) and y + (3)/(4) = 0`
`therefore` Focus is `((-17)/(32), (-3)/(4))`
Equation of directrix is `X = -a`
i.e., `x + (33)/(32) = (-1)/(2) or x = (-49)/(32)`