Question

**1. Derive the equation of parabola y = 4ax in standard form.
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Answer 1

Answer:

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

parabola

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 . 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.

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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, as shown in the diagram below

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 [from eq. (1)], 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.

Answer 2

Answer:

Parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane. The fixed line is called the directrix of the parabola and the fixed point F is called the focus. A line through the focus and perpendicular to the directrix is called the axis of the parabola. The point of intersection of parabola with the axis is called the vertex of the parabola. Let F be the focus and I the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X. By the definition of parabola, the mid point O is on the parabola and is called the vertex of the parabola. Take O as origin, OX the x-axis and OY perpendicular to it as the y - axis. Let the distance from the FM 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 as in Fig. Let P(x, y) be any point on the parabola such that PF = PB, where PB is perpendicular to l. The coordinates of B are (-a, y).

By the distance formula, we have PF = √((x - a)2 + y2) and PB = √((x + a)2) since PF = PB, we have √((x - a)2 + y2) = √((x + a)2)

i.e., (x - a)2 + y2 = (x + a)2

i.e., x2 - 2ax + a + y2 = x2 + 2ax + a2 i.e, y2 = 4ax (a>0). ∴ Any point on the parabola satisfies y2 = 4ax ------ (1) Conversely let P(x, y) satisfy (2)

∴ PF = √((x - a)2 + y2) = √((x - a)2 + 4ax) = √((x + a)2) = PB. -------- (2) and so P(x, y), lies on the parabola.

∴ From (1) and (2) we have proved that equation of parabola with vertex at the origin is y2 = 4ax.