Parabola whose Vertex at a given Point and Axis is Parallel to x-axis

We will discuss how to find the equation of the parabola whose vertex at a given point and axis is parallel to x-axis.

Let A (h, k) be the vertex of the parabola, AM is the axis of the parabola which is parallel to x-axis. The distance between the vertex and focus is AS = a and let P (x, y) be any point on the required parabola.

Now we shift the origin of co-ordinate system at A. Draw two mutually perpendicular straight lines AM and AN through the point A as x and y-axes respectively.

Parabola whose Vertex at a given Point and Axis is Parallel to x-axis

According to the new co-ordinate axes (x', y ') be the co-ordinates of P. Therefore, the equation of the parabola is (y')$^{2}$ = 4ax' (a > 0) …………….. (i)

Therefore, we get,

AM = x' and PM = y'

Also, OR = h, AR = k, OQ = x, PQ = y

Again, y = PQ

= PM + MQ

= PM + AR 

= y' + k              

Therefore, y' = y - k

And, x = OQ = OR + RQ

= OR + AM 

= h + x'         

Therefore, x' = x - h

Now putting the value of x' and y' in (i) we get

(y - k)$^{2}$ = 4a(x - h), which is the equation of the required parabola.

The equation (y - k)$^{2}$ = 4a(x - h) represents the equation of a parabola whose co-ordinate of the vertex is at (h, k), the co-ordinates of the focus are (a + h, k), the distance between its vertex and focus is a, the equation of directrix is x - h = - a or, x + a = h, the equation of the axis is y = k, the axis is parallel to positive x-axis, the length of its latus rectum = 4a, co-ordinates of the extremity of the latus rectum are (h  + a, k + 2a) and (h + a, k - 2a) and the equation of tangent at the vertex is x = h.

Solved example to find the equation of the parabola with its vertex at a given point and axis is parallel to x-axis:

Find the axis, co-ordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola y$^{2}$ + 4x + 2y - 11 = 0.

Solution:

The given parabola y$^{2}$ + 4x + 2y - 11 = 0.

y$^{2}$ + 4x + 2y - 11 = 0      

⇒ y$^{2}$ + 2y + 1 - 1 + 4x - 11 = 0

⇒ (y + 1)$^{2}$ = -4x + 12      

⇒ {y - (-1)}$^{2}$ = -4(x - 3)

⇒ {y - (-1)}$^{2}$ = 4 ∙ (-1) (x - 3) …………..(i)

Compare the above equation (i) with standard form of parabola (y - k)$^{2}$ = 4a(x - h), we get, h = 3, k = -1 and a = -1.

Therefore, the axis of the given parabola is along parallel to negative x-axis and its equation is y = - 1 i.e., y + 1 = 0.

The co-ordinates of its vertex are (h, k) i.e., (3, -1).

The co-ordinates of its focus are (h + a, k) i.e., (3 - 1, -1) i.e., (2, -1).

The length of its latus rectum = 4 units

The equation of its directrix is x + a = h i.e., x - 1 = 3 i.e., x - 1 - 3 = 0 i.e., x - 4 = 0.

● The Parabola

• Concept of Parabola
• Standard Equation of a Parabola
• Standard form of Parabola y22 = - 4ax
• Standard form of Parabola x22 = 4ay
• Standard form of Parabola x22 = -4ay
• Parabola whose Vertex at a given Point and Axis is Parallel to x-axis
• Parabola whose Vertex at a given Point and Axis is Parallel to y-axis
• Position of a Point with respect to a Parabola
  
• Parametric Equations of a Parabola
• Parabola Formulae
• Problems on Parabola

11 and 12 Grade Math 

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