The area of the region enclosed between the parabolas y2 = 2x – 1 and y2 = 4x – 3 is
- ⅓
- ⅙
- ⅔
- ¾
The Correct Option is A
Solution and Explanation
The correct option is(A): \(\frac{1}{3}\)
Area of the shaded region
\(=2∫_0^1(\frac{y^2+3}{4}-\frac{y^2+1}{2})dy\)
\(=2∫_0^1(\frac{1}{4}-\frac{y^2}{4})dy\)
\(=2[\frac{1}{4}-\frac{1}{12}]=\frac{1}{3}\)
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:
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Questions Asked in JEE Main exam
- If the components of \( \vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \) along and perpendicular to \( \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \) respectively are \( \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) \) and \( \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k}) \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:
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- Let \( A \) be a \( 3 \times 3 \) matrix such that \( X^T AX = 0 \) for all nonzero \( 3 \times 1 \) matrices \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \). \[ A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix} \] If \( \det(\text{adj}(2(A + I))) = 2^{\alpha} 3^{\beta} 5^{\gamma} \), \( \alpha, \beta, \gamma \in \mathbb{N} \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is:
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- Let \( f: [0, 3] \to A \) be defined by \( f(x) = 2x^3 - 15x^2 + 36x + 7 \) and \( g: [0, \infty) \to B \) be defined by \( g(x) = \frac{x}{x^{2025} + 1}. \) If both functions are onto and \( S = \{ x \in \mathbb{Z} : x \in A { or } x \in B \} \), then \( n(S) \) 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.