The area of the region enclosed by the parabolas \(y = 4 – x^2\) and \(3y = (x – 4)^2\) is in (sq. unit)?
\(\frac {14}{3}\)
\(4\)
\(\frac {32}{3}\)
\(6\)
The Correct Option is D
Solution and Explanation
The correct option is (D): \(6\).
Top Questions on Parabola
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 \( y^2 = 12x \) 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:
- 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:
- 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 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:
Questions Asked in JEE Main exam
Consider the following sequence of reactions :
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).- JEE Main - 2025
- Organic Chemistry
- A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is:
- JEE Main - 2025
- electrostatic potential and capacitance
- Three infinitely long wires with linear charge density \( \lambda \) are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?
- JEE Main - 2025
- electrostatic potential and capacitance
Two capacitors \( C_1 \) and \( C_2 \) are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are \( U_1 \) and \( U_2 \), respectively. Which of the given statements is true?
- JEE Main - 2025
- electrostatic potential and capacitance
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below:- JEE Main - 2025
- electrostatic potential and capacitance
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.