Find the coordinates of the focus, the axis of the parabola, the equation of directrix, and the length of the latus rectum for \(y^2 = 12x\).
Find the coordinates of the focus, the axis of the parabola, the equation of directrix, and the length of the latus rectum for \(y^2 = 12x\).
Solution and Explanation
The given equation is \(y^2= 12x. \)
Here, the coefficient of \(x\) is positive.
Hence, the parabola opens towards the right.
On comparing this equation with \(y^2 = 4ax\) , we obtain
\(4a= 12\)
\(⇒ a = 3\)\(\)
∴Coordinates of the focus \(= (a, 0) = (3, 0)\)
Since the given equation involves \(y^2 \), the axis of the parabola is the x-axis.
Equation of direcctrix ,\( x= -a \)
i.e., \(x = - 3\)
\(⇒ x+ 3 = 0 \)
Length of latus rectum : \( 4a= 4*3\) \(=12\)\(\) (Ans.)
Top Questions on Parabola
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Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
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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.