Find the coordinates of the focus, the axis of the parabola, the equation of directrix, and the length of the latus rectum for \(y^2 = - 8x\)
Find the coordinates of the focus, the axis of the parabola, the equation of directrix, and the length of the latus rectum for \(y^2 = - 8x\)
Solution and Explanation
The given equation is \(y^2= -8x\).
Here, the coefficient of x is negative. Hence, the parabola opens towards the left.
On comparing this equation with \(y^2= -4ax,\) we obtain
\(-4a= -8\)
\(⇒ a = 2\)
∴Coordinates of the focus = \((-a, 0) = (-2, 0)\) Since the given equation involves \(y^2\), the axis of the parabola is the x-axis.
Equation of directrix, \(x= a\)
i.e.\(,x = 2\)
The length of the latus rectum is \(4a= 8\) (Ans.)
Top Questions on Parabola
- If the line \( 3x - 2y + 12 = 0 \) intersects the parabola \( 4y = 3x^2 \) at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to:
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- Let \( P(4, 4\sqrt{3}) \) be a point on the parabola \( y^2 = 4ax \) and PQ be a focal chord of the parabola. If M and N are the foot of the perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:
- Let \( P(4, 4\sqrt{3}) \) be a point on the parabola \( y^2 = 4ax \) and PQ be a focal chord of the parabola. If M and N are the foot of the perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:
- Two parabolas have the same focus (4, 3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersect at points A and B, then \( (AB)^2 \) is equal to:
Notes on Parabola
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.