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Tricks & Tips of Parabola for NDA

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Term related to the parabola

(1) Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola. For the parabola `y^2=4ax`, `x`-axis is the axis.
Since equation has even power of `y` therefore the parabola is symmetric about `x`-axis i.e. about its axis.

(2) Vertex: The point of intersection of a parabola and its axis is called the vertex of the Parabola. For the parabola `y^2 = 4ax , 0(0, 0)` is the vertex.
The vertex is the middle point of the focus and the point of intersection of axis and directrix.

(3) Focal Distance : The distance of any point `P (x, y)` on the parabola from the focus is called the focal length (distance) of point `P`.
The focal distance of `P =` the perpendicular distance of the point `P` from the directrix.

(4) Double Ordinate: The chord which is perpendicular to the axis of Parabola or parallel to Directrix is called double ordinate of the Parabola.

(5) Focal Chord : Any chord of the parabola passing through the focus is called Focal chord.

(6) Latus Rectum: If a double ordinate passes through the focus of parabola then it is called as latus rectum. The extremities of the latus rectum are `L (a, 2a)` and `L'(a, - 2a)`. Since `LS = L'S = 2a`, therefore length of the latus rectum `LL' = 4a`.

(7) Parametric Equation of Parabola: The parametric equation of Parabola `y^2 = 4ax` are `x = at^2, y= 2at.`
Hence any point on this parabola is `(at^2, 2at)` which is also called as `'t'` point.
Q 1928612501

Find the equation of the parabola whose vertex
`(0, 0)`, passing through `(5, 2)` and symmetric
with respect to y-axis.
Class 11 Exercise 11.2 Q.No. 12
Solution:

Vertex `(0, 0)`, parabola passes through `(5, 2)`

symmetric ahout y-axis

Let the equation of parabola be `x^2 = 4ay`

`(5, 2)` lies on it

`:. 25 = 4a. 2 quad :. a = (25)/8`

`:.` Equation of parabola is

` x^2 = (25)/2 y` or `2x^2 = 25y`
Q 1977078886

Find the equation of parabola that satisfies the
given focus `(0, -3)` and directrix `y = 3`.
Class 11 Exercise 11.2 Q.No. 8
Solution:

Focus `(0, -3)` and directrix `y = 3`

vetiex is the mid point of `(0, -3), (0, 3)`

i.e.vertexis `(0 ,0)`

and `a = 3, quad :. 4a = 12`

`:.` Equation of parabolax `x^2 = -12y`
Q 2328280101

What is the focal distance of any point `P (x_1, y_1)`
'
on the parabola `y^2 = 4ax`?
NDA Paper 1 2011
(A)

`x_1 +y_1`

(B)

`x_1 + y_1`

(C)

`ax_1`

(D)

`a+ x_1`

Solution:

Given equation of parabola is, `y^2 = 4ax`

and `P(x_1 , y_1)` be a point on the parabola.

We know that.

`PS = e PM` (by definition of parabola)

`=> PS= PM` `( :. ` for parabola, `e = 1`)

`=> PS= ZN`

`=> PS = ZA+ AN`

`:. PS =a+ x_1`

which is the required focal distance.
Correct Answer is `=>` (D) `a+ x_1`
Q 2318280100

What is the area of the triangle formed by the
lines joining the vertex of the parabola `x^2 = 12 y` to
the end of the latusrectum?
NDA Paper 1 2011
(A)

`9` sq units

(B)

`12` sq units

(C)

`14` sq units

(D)

`18` sq units

Solution:

Equation of parabola is `x^2 = 12 y`. ................(i)

Length of focal distance `= 4a = 4 xx 3 = 12`

So, area of `Delta ABO` is

`=1/2 | (0,0,1), (6,3,1), (-6,3,1) |`

`= 1/2 (18+18) =1/2 xx 36 = 18` sq units
Correct Answer is `=>` (D) `18` sq units
Q 1977078886

Find the equation of parabola that satisfies the
given focus `(0, -3)` and directrix `y = 3`.
Class 11 Exercise 11.2 Q.No. 8
Solution:

Focus `(0, -3)` and directrix `y = 3`

vetiex is the mid point of `(0, -3), (0, 3)`

i.e.vertexis `(0 ,0)`

and `a = 3, quad :. 4a = 12`

`:.` Equation of parabolax `x^2 = -12y`

Different forms of standard parabola

Q 1658180004

The point on the parabola `y ^2 = 4ax` nearest to the

focus has its abscissa
NDA Paper 1 2015
(A)

`x = 0`

(B)

`x = a`

(C)

`x = a/2`

(D)

`x = 2a`

Solution:

:. Required abscissa is `x = 0`.
Correct Answer is `=>` (A) `x = 0`
Q 2318878709

The axis of the parabola `y^2 + 2x = 0` is
NDA Paper 1 2013
(A)

`x= 0`

(B)

`y =0`

(C)

`x =2`

(D)

`y =2`

Solution:

Given equation of parabola is

`y^2 + 2x= 0`

`:. y^2 = -2x`

which is of the form `y^2 = - 4ax`.

So, axis of the parabola is `y = 0`. i.e., `X`-axis.
Correct Answer is `=>` (B) `y =0`
Q 2368380205

The curve `y^2 = - 4ax (a > 0)` lies in
NDA Paper 1 2009
(A)

first and fourth quadrants

(B)

first and second quadrants

(C)

second and third quadrants

(D)

third and fourth quadrants

Solution:

Given, curve `y^2 = - 4ax`

It is clear from the figure that, the curve lies in the second and third quadrants.
Correct Answer is `=>` (C) second and third quadrants
Q 2409580418

The two parabolas `x^2 = 4y` and `y^2 = 4x` meet in two distinct points. One of these is the origin and the other is
BCECE Stage 1 2012
(A)

`(2, 2)`

(B)

`(4, - 4)`

(C)

`(4, 4)`

(D)

`(-2, 2)`

Solution:

Given, equations of parabola are `x^2= 4y` and `y^2 = 4x` ............(i)

`therefore (x^2/4)^2 = 4x`

`=> x^4-64x = 0`

`=> x = 0 , x = 4`
On putting the value of `x` in Eq. (i), we get `y = 0` and `y = 4, - 4` (`because y = - 4` does not satisfy the equation `x^2 = 4y`) Hence, points of intersection are `(0, 0)` and `(4, 4)`.
Correct Answer is `=>` (C) `(4, 4)`
Q 2435123962

The axis of the parabola
`9y^2 - 16x -12y- 57= 0` is
UPSEE 2012
(A)

`3y = 2`

(B)

`x+ 3y = 3`

(C)

`2x = 3`

(D)

`y = 3`

Solution:

Since, `9y^2 -16x -12y- 57= 0`

`=> (y- 2/3)^2 = 16/9 (x+ 61/16)`

Put `y-2/3 =y` and `x+ 61/16 = x`

`=> y^2 = 4 (4/9) x`

Axis of this parabola is `y = 0`

`=> y -2/3 = 0 => y = 2/3`

`=> 3y = 2`
Correct Answer is `=>` (A) `3y = 2`

Finding equation of parabola

Finding equation of parabola
Q 1739101012

Find the equation of each of the following parabolas
(i) directrix `= 0`, focus at `(6, 0)`
(ii) vertex at `(0, 4)`, focus at `(0, 2)`
(iii) focus at `( -1,- 2),` directrix `x - 2y + 3 = 0`
NCERT Exemplar
Solution:

(i) Given that, directrix `= 0` and focus `= (6, 0)`

So, the equation of the parabola

` (x- 6)^2 + y^2 = x^2`

` => x^2 + 36 - 12x + y^2 = x^2`

` => y^2 - 12x + 36 = 0`

(ii) Given that, vertex `= (0, 4)` and focus `= (0, 2)`

So. the equation of parabola is

` sqrt( (x - 0)^2 + (y - 2)^2 = | y - 6 |`

` => x^2 + y^2 - 4y+ 4 = y^2 - 12y + 36`

`=> x^2 - 4y + 12y - 32 = 0`

` => x^2 + 8y - 32 = 0`

` => x^2 = 32 - 8y`

(iii) Given that, focus at `(-1.- 2)` and directrix `x - 2y + 3 = 0`

So, the equation ot' parabola is `sqrt( (x + 1)^2 + ( y + 2)^2) = | ( x - 2y + 3)/sqrt( 1 + 4) |`

`=> x^2 + 2x + 1 + y^2 + 4y + 4 = 1/5 [x^2 + 4y^2 + 9 - 4xy -12y + 6x]`

`=> 4x^2 + 4xy+ y^2 +4x + 32y + 16 = 0`
Q 1733556442

What is the equation of parabola whose vertex is at
`(0, 0)` and focus is at `(0,- 2)?`
NDA Paper 1 2014
(A)

`y^ 2 + 8x = 0`

(B)

`y^ 2 - 8x = 0`

(C)

`x^ 2 + 8y = 0`

(D)

`x^ 2 - 8y = 0`

Solution:

Given vertex of the parabola `= (0, 0)`

and focus of the parabola `= (0, - 2)`

Let `P` be any point on the parabola, then equation directrix is

`y - 2 = 0`

:. Equation of parabola is

` => (PS =PM)`

`sqrt( (x-0)^2 + (y+2)^2) = (|y-2|)/sqrt(1)`

` => ( sqrt(x^2 + (y + 2 )^2 )^2) = |y-2|^2`

` => x^2 + y^2 + 4 + 4y = y^2 + 4 - 4y`

` x^2 = - 8y`, which is the required equation of parabola
Correct Answer is `=>` (C) `x^ 2 + 8y = 0`
Q 2359001814

What is the equation to the parabola, whose
vertex and focus are on the `X`-axis at distances `a`
and `b` from the origin respectively? `(b >a > 0)`
NDA Paper 1 2007
(A)

`y^2 = 8(b - a)(x - a)`

(B)

`y^2 = 4(b +a)(x - a)`

(C)

`y^2 = 4 (b - a)(x + a)`

(D)

`y^2 = 4(b - a)(x - a)`

Solution:

Since, the focus and vertex of the parabola are on
`X`-axis, therefore, its direction is parallel to `X`-axis and axis of
parabola is `X`-axis. Let the equation of the directrix be `x = k`. The
directrix meets the axis of parabola at `(k, 0)`. But vertex is
mid-point of the line segment joining the focus to the point, where
directrix meets the axis of the parabola

`:. (k+b)/2 =a => k = 2a -b`

So, equation of the directrix is `x = 2a - b`.
Let `(x, y)` be a point on the parabola, then

`(x-b)^2 +y^2 = |(x-2a+b)/1 |^2`

`=> x^2 + b^2 - 2bx + y^2 = x^2 + 4a^2 + b^2 - 4ax - 4ab + 2bx`

`=> y^2 = 4bx - 4ax - 4ab + 4a^2`

`=> 4x(b -a)- 4a(b- a)`

`:. y^2 = 4(b - a)(x- a)`

which is the required equation of parabola.


`text (Alternate Method)`

Graph of the given condition of parabola.


`:. OA =a, OS=b`

`:. AS = OS -OA`

`= b -a`

Hence, equation of parabola whose axis is `X`-axis and having
vertex `(a, 0)` and length of foci is `4AS = 4(b- a)`, is

`(y-0)^2 = 4(b-a)(x-a)`

`=> y^2 = 4(b-a)(x-a)`
Correct Answer is `=>` (D) `y^2 = 4(b - a)(x - a)`
Q 1779234116

If the vertex of the parabola is the point `(- 3, 0)` and the directrix is the
line `x + 5 = 0`, then its equation is
NCERT Exemplar
(A)

`y^2 = 8 (x + 3)`

(B)

`x^2 = 8 (y + 3)`

(C)

`y^2 = - 8 (x + 3)`

(D)

`y^2 = 8 (x + 5)`

Solution:

Here, vertex `= (- 3, 0)`

`:. a = - 3` and directrix, `x + 5 = 0`

Since, axis of the parabola is a line perpendicular to directrix and `A` is the mid-point of
AS.

Then, `-3 = (x_1 - 5)/2`

`=> - 6 = x_1 - 5 => x_1 = - 1`,

` 0 = ( 0 + y_1)/2 => y_1 = 0`

` :. S = (-1, 0)`

` ∵ PM = PS`

` => | x + 5| = sqrt((x + 1)^2 + y^2)`

` => x^2 + 2x + 1 + y^2 = x^2 + 10x + 25`

` => y^2 = + 8x + 24`

` => y^2 = + 8 (x + 3)`
Correct Answer is `=>` (A) `y^2 = 8 (x + 3)`
 
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