Since e = 1 , for a parabola, we note that the parabola is the locus of points in a plane that are equidistant from both the directrix and the focus.

**Parabola**

Since *e *= 1 , for a parabola, we note that the parabola is
the locus of points in a plane that are equidistant from both the directrix and
the focus.

Let *S *be the focus and *l *be the directrix.

Draw *SZ *perpendicular to the line *l .*

Let us assume *SZ *produced as *x *-axis and the perpendicular bisector
of *SZ *produced as *y *- axis. The intersection of this
perpendicular bisector with *SZ *be the origin *O *.

Let *SZ *= 2*a *. Then *S *is (*a*, 0) and
the equation of the directrix is *x *+ *a *= 0 .

Let *P*(*x*, *y*) be the moving point in the
locus that yield a parabola. Draw *PM *perpendicular to the directrix. By definition, *e *= *SP*/*PM *= 1. So, *SP*^{2}
= *PM*^{2} .

Then, (*x *- *a*)^{2} + *y*^{2} =
(*x *+ *a*)^{2}. On simplifying, we get *y*^{2} = 4*ax *which is the equation
of the **parabola in the
standard form****.**

The other standard forms of parabola are *y*^{2} =
-4*ax*, *x*^{2} = 4*ay *, and *x*^{2} = -4*ay
*.

·
The line perpendicular to the directrix and passing through the
focus is known as the **Axis** of
the parabola.

·
The intersection point of the axis with the curve is called **vertex **of the parabola

·
Any chord of the parabola, through its focus is called **focal chord **of the parabola

·
The length of the focal chord perpendicular to the axis is
called **latus
rectum **of
the parabola

Find the length of Latus rectum of the parabola *y*^{2}
= 4*ax *.

Equation of the parabola is *y*^{2} = 4*ax*.

Latus rectum *LL*¢ passes through the focus (*a*, 0) . Refer (Fig.5.18)

Hence the point *L *is (*a*, *y*_{1}) .

Therefore *y*_{1}^{2}
= 4*a*^{2}.

Hence *y*_{1} = ± 2*a *.

The end points of latus rectum are (*a*, 2*a*) and (*a*,
-2*a*) .

**Therefore length of
the latus rectum ***LL***¢**** = 4 a **

**Note**

The standard form of the parabola *y*^{2} = 4*ax *has
for its vertex (0, 0) , axis as *x *-axis, focus as (*a*, 0) . The parabola *y*^{2}
= 4*ax *lies completely on the non-negative side of the *x-*axis.
Replacing y by –*y *in *y*^{2}
= 4*ax, *the equation remains the same. so the parabola *y*^{2}
= 4*ax *is symmetric about *x*-axis; that is, *x*-axis is the axis and symmetry of *y*^{2}
= 4*ax*

When the vertex is (*h*, *k*) and the axis of symmetry
is parallel to *x *-axis, the equation of the parabola is either ( *y *- *k *)^{2}
= 4*a*(*x *- *h*) or ( *y *- *k *)^{2} = - 4*a*(*x
*- *h*) (Fig. 5.19, 5.20).

When the vertex is (*h*, *k*) and the axis of symmetry
is parallel to *y *-axis, the equation of the parabola is either (*x *- *h*)^{2}
= 4*a*( *y *- *k *) or (*x *- *h*)^{2} = - 4*a*(
*y *- *k *) (Fig. 5.21, 5.22).

Tags : Equation, Definition, Example, Solution, vertex , 12th Mathematics : UNIT 5 : Two Dimensional Analytical Geometry II

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12th Mathematics : UNIT 5 : Two Dimensional Analytical Geometry II : Parabola | Equation, Definition, Example, Solution, vertex

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