Question:
The focus of the parabola $( y + 1)^2 = -8(x +2)$ is
The focus of the parabola $( y + 1)^2 = -8(x +2)$ is
Updated On: Jun 7, 2024
- (-4, -1)
- (-1, -4)
- (1, 4)
- (4, 1)
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The Correct Option is A
Solution and Explanation
Given, $(y+1)^{2}=-8(x+2)$
$Y^{2}=-8 X$ ...(i)
Here, $Y=y+1, \, X=(x+2)$
Vertices $(X=0, Y=0)=(-2,-1)$
Comparing E (i) from $y^{2}=4 ax$
$4 a=-8$
$a=-2$
Focus $=(-2-2,-1)$
$=(-4,-1)$
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.