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Parabola

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Types of parabola

x + a = 0 y² = 4ax, a > 0 Y² = - 4ax, a > 0

s

Transformation of parabola.

(a) (y - R)² = 4a (x - h) , a > 0

The vcetex will be (h , k), openitng of parabola will be on = ve side of axis, axis will be || to x axis and dire ctrox well be || to y axis.

(b) (y - R²) = 4a (x-h), a < 0

Same as above except, the opening of parabolanwill be on -ve side of x - axis .

s

(c) (x -h ²) = 4a (y - yR), a > 0

The vertex will be at (h,K) opening of parabola will be on +ve side of yaxis , axis will be || t y - axid and directrix || to x - axis

(d) (x -h ²) = 4a (y-K), a < 0

Same ad above except, lthe opening of parabola will be on -ve side of y-axis .

Illustration 2.

Find the vertex, axis, focus, and latus rectum of parabola

4y²+ 12x -x - 20y + 67 = 0

Ans The equation can be written as

y²- 5y = - 3x - 67/4 .

i.e. (y-5/2²) = - 3x - 64/2 + 25/4 .

= - 3(x + 7/2)

So, this a transformed parabola whose vertex is (-7/2, 5/2), the axis is y = 5/2.

The length of latus rectum = |4a| = 3

and a = -3/4 .

So, focis is (-7/2 - 3/4 1 5/52)

= (-17/4 1 5/2).

Natations :

For standard parabola (y² = 4ax)

1) S

y²- 4ax

2) S₁

y ₁²- 4ax₁

3) T

yy₁-2a(x + x₁)

4) F

(at ,²2at)

Position of a point (x₁,y₁) w. r. t y² = 4ax

If S₁ > 0 => Outside parabola.

S₁ < 0 => Inside parabola.

S₁ 0 => On parabola.

Why ?

Suppose a point is outside parabola .

s So, PL > PM.

=> PL ² PM²

or y₁² > y₂²

or y₁² > 4ax₁(

M lics oh parabola).

So, S₁ > 0 => Point is outside parabola .

Similarily when point is inside the parabola .

S₁ < 0 .

Dumb Question:- What dose inside and outside the parabola mean in a curre like parabola which is not closed ?

Ans :- the wird "outside" refers to the region from where tangent can be drawn . On other hand, the region from where tangent cannot be drawn is fefference as "in side" the parabola.

Par ame tric form.

x = at², y = 2at where t is a parameter represents the parametric form.

(at², 2at) is general point on parabola y² = 4ax .

Illustration 2.

Find the equations of the parabola if the extremeties of its latus ractum are (3,5) and (3,7).

Ans.

Now the length of latus ractuin is

So, 4a = 2 or a = 1/2 .

Now middle point of catus ractum is

(3+3/2, 5+7/2) = (3,6) which is focus of parabola.

So the two cases as shown in figure below are possible.

Since the vertex is at a distance a away from the parabola the vertices are (7/2, 6) and (5/2,6).

Now lBy transformation of parabolas the two parabolas possible are

(y-6)² = 4(1/2) (x-5/2)

& (y-6)² = 4(1/2) (x-7/2).

So, equation of parabolas are

(y-6)² = 2 (x-5/2)

& (y-6)² =2 (x-7/2)

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