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

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

Let A (h, k) be the vertex of the parabola, AM is the axis of the parabola which is parallel to y-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 y and x-axes respectively.

According to the new co-ordinate axes (x', y ') be the co-ordinates 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 co-ordinate of the vertex is at (h, k), the co-ordinates 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 y-axis, the length of its latus rectum = 4a, co-ordinates 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 y-axis:

Find the axis, co-ordinates 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 y-axis and its equation is x = h i.e., x = 3 i.e., x - 3 = 0.

The co-ordinates of its vertex are (h, k) i.e., (3, 2).

The co-ordinates 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.