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
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 CBSE Class XI exam
Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
- CBSE Class XI
- Stress-strain curve
- The reaction CH3CH2I + KOH(aq) \(\rightarrow\) CH3CH2OH + KI is classified as
- CBSE Class XI
- Organic Chemistry - Some Basic Principles and Techniques
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.- CBSE Class XI
- Discovering Tut: The saga Continues
Draw the Lewis structures for the following molecules and ions: \(H_2S\), \(SiCl_4\), \(BeF_2\), \(CO_3^{2-}\) , \(HCOOH\)
- CBSE Class XI
- Kossel-Lewis Approach to Chemical Bonding
- The following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant
λ (nm) 500 450 400 v × 10–5(cm s–1) 2.55 4.35 5.35 - CBSE Class XI
- Atomic Models
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.