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
- Let A and B be the two points of intersection of the line \( y + 5 = 0 \) and the mirror image of the parabola \( y^2 = 4x \) with respect to the line \( x + y + 4 = 0 \). If \( d \) denotes the distance between A and B, and \( a \) denotes the area of \( \Delta SAB \), where \( S \) is the focus of the parabola \( y^2 = 4x \), then the value of \( (a + d) \) is:
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 the points $A$ and $B$, then $(AB)^2$ 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:
- 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:
- Let the point $ P $ of the focal chord $ PQ $ of the parabola $ y^2 = 16x $ be $ (1, -4) $. If the focus of the parabola divides the chord $ PQ $ in the ratio $ m : n $, gcd($m, n$) = 1, then $ m^2 + n^2 $ is equal to:
Questions Asked in CBSE Class XI exam
- We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
- CBSE Class XI
- Motion in a straight line
- What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH- CBSE Class XI
- Oxidation Number
- Using s, p, d notations, describe the orbital with the following quantum numbers.
- n=1, l=0
- n = 3, l=1
- n = 4, l =2
- n=4, l=3
- CBSE Class XI
- Developments Leading to the Bohr’s Model of Atom
- Write the following as intervals:
(i) {\(x: x ∈ R, -4 < x ≤ 6\)}
(ii) {\(x: x ∈ R, -12 < x < -10\)}
(iii) {\(x: x ∈ R, 0 ≤ x < 7\)}
(iv) {\(x: x ∈ R, 3 ≤ x ≤ 4\)}- CBSE Class XI
- Sets and their Representations
- Reduce the following equations into intercept form and find their intercepts on the axes.
(i) \(3x+2y-12=0\)
(ii) \(4x-3y=6\)
(iii) \(3y+2=0. \)- CBSE Class XI
- Distance of a Point From a Line
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.