We will discuss how to find the equation of the parabola whose vertex at a given point and axis is parallel to xaxis.
Let A (h, k) be the vertex of the parabola, AM is the axis of the parabola which is parallel to xaxis. 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 coordinate system at A. Draw two mutually perpendicular straight lines AM and AN through the point A as x and yaxes respectively.
According to the new coordinate axes (x', y ') be the coordinates 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 coordinate of the vertex is at (h, k), the coordinates 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 xaxis, the length of its latus rectum = 4a, coordinates 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 xaxis:
Find the axis, coordinates 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 xaxis and its equation is y =  1 i.e., y + 1 = 0.
The coordinates of its vertex are (h, k) i.e., (3, 1).
The coordinates 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.
`11 and 12 Grade Math
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