Find the equation of the parabola that satisfies the following conditions: Vertex \((0, 0) \)passing through \((2, 3) \)and the axis is along the x-axis
Find the equation of the parabola that satisfies the following conditions: Vertex \((0, 0) \)passing through \((2, 3) \)and the axis is along the x-axis
Solution and Explanation
Given that
vertex is \((0, 0) \)and the axis of the parabola is the x-axis,
then the equation of the parabola is either of the forms \(y^2= 4ax \) or \(y^2= -4ax. \)
The parabola passes through the point \((2, 3),\) which lies in the first quadrant.
Therefore, the equation of the parabola is of the form \(y^2= 4ax\)
, while points \((2, 3)\) must satisfy the equation \(y^2= 4ax.\)
\(∴ 3^2 = 4a(2)\)
\(3^2 = 8a\)
\(9 = 8a\)
\(a = \dfrac{9}{8}\)
Thus, the equation of the parabola is
\(y^2 = 4 (\dfrac{9}{8})x\)
\(⇒y^2=\dfrac{9x}{2}\)
\(⇒2y^2 = 9x\)
∴ The equation of the parabola is \(2y^2 = 9x.\) (Ans)
Top Questions on Parabola
- An equilateral triangle OAB is inscribed in the parabola $y^2 = 4x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is
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:
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16
- 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 the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
Questions Asked in CBSE Class XI exam
- Balance the following redox reactions by ion-electron method:
(a)MnO4- (aq) + I - (aq) → MnO2(s) + I2(s) (in basic medium)
(b) MnO4- (aq) + SO2(g) → Mn2+(aq) +HSO4- (aq) (in acidic solution)
(c) H2O2(aq)+Fe2+(aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O72-+ SO2(g) → Cr3+ (aq) +SO42- (aq) (in acidic solution)- CBSE Class XI
- Oxidation Number
- Write the resonance structures for SO3 , NO2 and NO3-
- CBSE Class XI
- Kossel-Lewis Approach to Chemical Bonding
- At 700 K, equilibrium constant for the reaction:
\(H_2 (g) + I_2 (g) ⇋ 2HI (g)\)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?- CBSE Class XI
- Law Of Chemical Equilibrium And Equilibrium Constant
- Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17.
- CBSE Class XI
- Statistics
Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
- CBSE Class XI
- Statistics
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.