We will discuss how to find the equation of the parabola whose vertex at a given point and axis is parallel to yaxis.
Let A (h, k) be the vertex of the parabola, AM is the axis of the parabola which is parallel to yaxis. 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 y and xaxes respectively.
According to the new coordinate axes (x', y ') be the coordinates of P. Therefore, the equation of the parabola is (x’)\(^{2}\) = 4ay' (a > 0) …………….. (i)
Therefore, we get,
AM = y' and PM = x'
Also, OR = k, AR = h, OQ = y, PQ = x
Again, x = PQ
= PM + MQ
= PM + AR
= x' + h
Therefore, x' = x  h
And, y = OQ = OR + RQ
= OR + AM
= k + y'
Therefore, y' = y  k
Now putting the value of x' and y' in (i) we get
(x  h)\(^{2}\) = 4a(y  k), which is the equation of the required parabola.
The equation (x  h)\(^{2}\) = 4a(y  k) represents the equation of a parabola whose coordinate of the vertex is at (h, k), the coordinates of the focus are (h, a + k), the distance between its vertex and focus is a, the equation of directrix is y  k =  a or, y + a = k, the equation of the axis is x = h, the axis is parallel to positive yaxis, the length of its latus rectum = 4a, coordinates of the extremity of the latus rectum are (h + 2a, k + a) and (h  2a, k + a) and the equation of tangent at the vertex is y = k.
Solved example to find the equation of the parabola with its vertex at a given point and axis is parallel to yaxis:
Find the axis, coordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola x\(^{2}\)  y = 6x  11.
Solution:
The given parabola x\(^{2}\)  y = 6x  11.
⇒ x\(^{2}\)  6x = y  11.
⇒ x\(^{2}\)  6x + 9 = y  11 + 9
⇒ (x  3)\(^{2}\) = y  2
⇒ (x  3)\(^{2}\) = 4 ∙ ¼(y  2) ………….. (i)
Compare the above equation (i) with standard form of parabola (x  h)\(^{2}\) = 4a(y  k), we get, h = 3, k = 2 and a = ¼.
Therefore, the axis of the given parabola is along parallel to positive yaxis and its equation is x = h i.e., x = 3 i.e., x  3 = 0.
The coordinates of its vertex are (h, k) i.e., (3, 2).
The coordinates of its focus are (h, a + k) i.e., (3, ¼ + 2) i.e., (3, \(\frac{9}{4}\)).
The length of its latus rectum = 4a = 4 ∙ ¼ = 1 unit
The equation of its directrix is y + a = k i.e., y + ¼ = 2 i.e., y + ¼  2 = 0 i.e., y  \(\frac{7}{4}\) = 0 i.e., 4y  7 = 0.
`11 and 12 Grade Math
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