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previous year question of Parabola for NDA

Previous Year Parabola Question For NDA

Previous Year Parabola Question For NDA
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 1619267110

The line `2y = 3x + 12` cuts the
parabola `4 y = 3x^2`

Where does the line cut the parabola?
NDA Paper 1 2014
(A)

At `(-2, 3)` only

(B)

At `(4, 12)` only

(C)

At both `(-2, 3)` and `(4, 12)`

(D)

Neither at `(-2, 3)` nor at `(4, 12)`

Solution:

Equation of line is `2y = 3x + 12 ... (i)`

Equation of parabola is

`4y = 3x^2... (ii)`

On putting the value of y from Eq. (i) to Eq. (ii), we get

`2(3x + 12) = 3x^2`

`=> 3x^2 - 6x- 24 = 0 => x^2 - 2x- 8 = 0`

`=> (x- 4)(x + 2) = 0 => x = 4` and `x = -2`

On putting these values in Eq. (ii), we get

`y = 12` and `y = 3`

Hence, points are `(-2, 3)` and `(4, 12)`.
Correct Answer is `=>` (C) At both `(-2, 3)` and `(4, 12)`
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 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 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 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 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 2348480303

Consider the parabolas `S_1 equiv y^2 - 4ax =0` and

`S_2 equiv y^2 - 4bx = 0`, `S_2` will contain `S_1` if
NDA Paper 1 2008
(A)

`a > b > 0`

(B)

`b > a > 0`

(C)

`a >0, b < 0` but `| b | > a`

(D)

`a < 0, b > 0` but `b > | a |`

Solution:

If `a, b > 0`, then

`S_2` will contain `S_1`, if

Latusrectum of `S_2 >` Latusrectum of `S_1`

`=> 4b > 4a => b > a`

`:. b > a > 0`
Correct Answer is `=>` (B) `b > a > 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)`
 
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