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
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 P be the parabola, whose focus is $ (-2, 1) $ and directrix is $ 2x + y + 2 = 0 $. Then the sum of the ordinates of the points on P, whose abscissa is -2, is:
- 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:
Questions Asked in CBSE Class XI exam
- Find the derivative of
(i) 2x - \(\frac{3}{4}\)
(ii)(5x3+3x-1)(x-1)
(iii) x-3(5+3x) (iv)x5(3-6x-9)
(v) x-4(3-4x-5) (vi)\(\frac{2}{x+1}-\frac{x^2}{3x-1}\)- CBSE Class XI
- Derivatives
- Find the direction in which a straight line must be drawn through the point (-1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point
- CBSE Class XI
- Straight lines
- The following three compound words end in -ship. What does each of them mean?
airship flagship lightship - CBSE Class XI
- We're not afraid to Die if we can all be together
- Write the resonance structures for SO3 , NO2 and NO3-
- CBSE Class XI
- Kossel-Lewis Approach to Chemical Bonding
- Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :
Temperature Pressure thermometer A Pressure thermometer B Triple-point of water 1.250 × 10\(^5\) Pa 0.200 × 10\(^5\) Pa Normal melting point of sulphur 1.797× 10\(^5\) Pa 0.287 × 10\(^5\) Pa - What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
- What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
- CBSE Class XI
- Thermal Expansion
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.