Find the standard equation of a parabola
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Answer:
Equation of Parabola
The set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the fixed line) in the plane is a parabola. The fixed line is the ‘directrix’ and the fixed point is the ‘focus’ . In simple words, it is the shape described when we throw a ball in the air.
Further, a line through the focus and perpendicular to the directrix is the axis of the parabola. Also, the point of intersection of the parabola with the axis is the vertex.
Standard Equations of the Parabola
When the vertex of a parabola is at the ‘origin’ and the axis of symmetry is along the x or y-axis, then the equation of the parabola is the simplest. Here is a quick look at four such possible orientations:
parabola
Of these, let’s derive the equation for the parabola shown in . As can be seen in the diagram, the parabola has focus at (a, 0) with a > 0. Also, the directrix x = – a. Let F be the focus and l, the directrix. Also, let FM be perpendicular to the directrix and bisect FM at point O. Produce MO to X.
By the definition of the parabola, the mid-point O is on the parabola and is called the vertex of the parabola. Next, take O as origin, OX the x-axis and OY perpendicular to it as the y-axis. Let the distance from the directrix to the focus be 2a. Then, the coordinates of the focus are: (a, 0), and the equation of the directrix is x + a = 0 .
Derivation of the Equation
Now, we take a point P(x, y) on the parabola such that, PF = PB … (1) … where PB is perpendicular to l. The coordinates of the point B are (- a, y). Also, by the distance formula, we know that
PF = √ {(x – a)2 + y2}
Also, PB = √ (x + a)2
Since, PF = PB , we get
√ {(x – a)2 + y2} = √ (x + a)2
So, by squaring both sides, we have (x – a)2 + y2 = (x + a)2 or, x2 – 2ax + a2 + y2 = x2 + 2ax + a2 . Further, by the solving the equation, we have y2 = 4ax … where a > 0. Therefore, we can say that any point on the parabola satisfies the equation:
y2 = 4ax … (2)
Let’s look at the converse situation now. If P(x, y) satisfies equation (2), then
PF = √ {(x – a)2 + y2}
= √ {(x – a)2 + 4ax} … using the RHS of equation (2)
= √ (x2 – 2ax + a2 + 4ax) = √ (x2 + 2ax + a2)
= √ (x + a)2 = PB … (3)
Hence, we conclude that the point P(x, y) satisfying eq. (2) lies on the parabola. Also, equations (2) and (3) prove that the equation to the parabola with vertex at the origin, focus at (a, 0) and directrix x = – a, is y2 = 4ax.
Important
Since a > 0, x can assume any positive value or zero but no negative value. Hence, the curve extends indefinitely far into the first and the fourth quadrants. The axis of the parabola is the positive x-axis.
Similarly, we can derive the equations of the parabolas in the other three figure as:
(b): y2 = – 4ax,
(c): x2 = 4ay,
(d): x2 = – 4ay.
These four equations are the Standard Equations of Parabolas.
Observations
A parabola is symmetric with its axis. If the equation has a y2 term, then the axis of symmetry is along the x-axis and if the equation has an x2 term, then the axis of symmetry is along the y-axis.
For axis of symmetry along the x-axis:
It opens to the right if the coefficient of x is positive
It opens to the left if the coefficient of x is negative
For axis of symmetry along the y-axis:
It opens upwards if the coefficient of y is positive
It opens downwards if the coefficient of y is negative
Latus Rectum
Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose endpoints lie.
Let’s find the length of the latus rectum of the parabola y2 =4ax
By definition of the parabola; AF = AC. However, it can be seen in the diagram above that AC = FM (since both AC and FM are perpendicular to the directrix and AF is perpendicular to the x-axis). Also, FM = 2a. Hence, we can conclude that AF = 2a. Further, the figure is symmetric with respect to the x-axis. Hence, the length of the latus rectum,
AB = AF + FB = 2a + 2a = 4a.
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