Mathematics Must Do Problems of Ellipse for NDA

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Must Do Problems of Ellipse for NDA

Must Do Problem of ellpise

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Q 2816401379

Consider the equation of
ellipse
` 12x^2 + 4y^2 +24x - 16y + 25 = 0`
The centre of the ellipse is

(A)

(4,-1)

(B)

(2,1)

(C)

(-1,2 )

(D)

(-3,2)

Solution:

Coordinates of centre of the ellipse
are given by

`x + 1 = 0 `and `y - 2 = 0`

`=> x = - 1` and `y = 2`

`:.` Centre of the ellipse is `(- 1, 2)`.

Correct Answer is `=>` (C) (-1,2 )

Q 2816401379

Consider the equation of
ellipse
` 12x^2 + 4y^2 +24x - 16y + 25 = 0`
The length of major and minor axes are

(A)

`2,4`

(B)

`sqrt 3 ,1`

(C)

`2, 2 sqrt 3`

(D)

`4, 6 `

Solution:

Let a and b the length of the
semi-minor and semi-major axes, then

`a^2 = 1/4` and `b^2 = 3/4`

`:.` Length of major axis

`= 2b = 2 xx sqrt ( 3/4) = sqrt 3`

Length of minor axis=` 2a = 2 xx sqrt (1/4) = 1`

Correct Answer is `=>` (B) `sqrt 3 ,1`

Q 2816401379

Consider the equation of
ellipse
` 12x^2 + 4y^2 +24x - 16y + 25 = 0`
The eccentricity of the ellipse is

(A)

`sqrt (1/3)`

(B)

`3/4`

(C)

`sqrt (2/3 )`

(D)

`1/4`

Solution:

Eccentricity `= sqrt (1- a^2/b^2) = sqrt (1- (1/4)/(3/4) )`

`= sqrt (1-1/3) = sqrt (2/3)`

Correct Answer is `=>` (C) `sqrt (2/3 )`

Q 2816401379

Consider the equation of
ellipse
` 12x^2 + 4y^2 +24x - 16y + 25 = 0`
Coordinates of the foci are

(A)

`( 0 ,1 pm 1/sqrt3 )`

(B)

`( 0, 2 pm sqrt (1/3) )`

(C)

`(1, 2 pm 1/(sqrt 2 ) )`

(D)

`(-1 , 2 pm 1/(sqrt 2))`

Solution:

We have, `be = (sqrt 3)/2 xx sqrt (2/3) = 1/(sqrt 2)`

Coordinates of foci are given by

`x + 1 = 0, y - 2 = pm be`

Thus, foci are ` (-1,2 pm 1/(sqrt 2))`

Correct Answer is `=>` (D) `(-1 , 2 pm 1/(sqrt 2))`

Q 2855191964

If a hyperbola passes through the foci of the ellipse `x^2/25 + y^2/16 = 1` and its

transverse and conjugate axis coincide with major
and minor axes of the ellipse and product of the
eccentricities is 1, then
The equation of hyperbola is

(A)

`x^2/9 - y^2/16 =1`

(B)

`x^2/9- y^2/25 = 1`

(C)

`-x^2/9 + y^2/25 = 1`

(D)

`x^2/16 - y^2/25 =1`

Solution:

Eccentricity of ellipse

`= sqrt( (25 -16 )/25) = 3/5`

`:.` Eccentricity of hyperbola `= 5/3`

Foci of ellipse `( pm 3 ,0)`.

`:.` Equation of hyperbola `x^2/9^2 - y^2/b^2 = 1`

where, `b^2 = 9 (e^2 -1 )`

`= 9 (25/9 -1) = 16`

`:.` Equation of hyperbola is `x^2/9 - y^2/16 = 1`

Correct Answer is `=>` (A) `x^2/9 - y^2/16 =1`

Q 2855191964

(A)

`(5,0)`

(B)

`(5 sqrt 3 ,0 )`

(C)

`(0, 5 )`

(D)

`( 0 , 5 sqrt 3 )`

Solution:

Focus of hyperbola is

`( pm 3 xx 5/3 ,0 ) = ( pm 5, 0 )`

Correct Answer is `=>` (A) `(5,0)`

Q 2825091861

Consider the following statements
I. The area of the ellipse `2x^2 + 3 y^2 = 6` is more than
the area of the circle `x^2 + y^2 - 2x+ 4y + 4 = 0`.

II. The equation `3x^2 + 4y^2 -18x + 16 y + 43 = k`
represents an ellipse if `k < 0`.

Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

None of these

Solution:

I. Given ellipse is `x^2/3 + y^2/2 = 1`,

whose area is `= pi sqrt 3 * sqrt 2 = pi sqrt 6`

Circle is `x^2 + y^2 - 2x + 4y + 4 = 0`

or `(x-1)^2 + (y-2)^2 = 1`

Its radius is 1. Hence, area is `pi`.

So, Statement I is true.

II. The given equation can be written as

`3(x-3)^2 + 4 (y+2)^2 = k`

So, no locus for `k < 0`

Hence, Statement ll is false.

Correct Answer is `=>` (A) Only I

Q 2815591469

If the foci of the ellipse `x^2/16 + y^2/b^2 =1` and the
hyperbola `x^2/144 - y^2/81 = 1/25` coincide, then the value of
`b^2` is

(A)

(B)

(C)

(D)

Solution:

For hyperbola,

`e^2 = 1+ b^2/a^2 =1 + 81/144 = 225/144`

`:. e =15/12 = 5/4 ` , i.e., `e > 1`

Also, `a^2 = 144/25`

Hence, the foci are `(pm ae,0 )`

i.e., `(pm 12/5 * 5/4 , 0 ) = ( pm 3, 0)`

Now, the f(Ki coincide, therefore for ellipse

`ae = 3` or `a^2 e^2 = 9 => a^2 (1- b^2/a^2) = 9`

`=> a^2 - b^2 = 9 => 16 - 9 = b^2`

`:. b^2 = 7`

Correct Answer is `=>` (C) 7

Q 2815491369

A man running around a race course notes that
the sum of the distances of two flag posts from
him is always 10 m and the distance between the
flag postc is 8 m. The area of the path, he
encloses in square metres is

(A)

`15 pi`

(B)

`12 pi`

(C)

`18 pi`

(D)

`8 pi`

Solution:

Clearly, the race course will be an
ellipse with the flag posts as its foci. If a
and b are the semi-major and
semi-minor a.xcs of the ellipse, then
`2a = 10 `and `2ae = 8`

`:. a = 5 , e = 4/5 ` and `b^2 = a^2 (1- e^2) = 9`

`:.` Area of the ellipse

`= pi a b = pi * 5 * 3 = 15 pi` sq m

Correct Answer is `=>` (A) `15 pi`

Q 2845491363

A circle is drawn with the two foci of an ellipse

`x^2/a^2 + y^2/b^2 =1` at the end of the diameter. What is

the equation of the circle?

(A)

`x^2 + y^2 = a^2 + b^2`

(B)

`x^2 + y^2 = a^2 -b^2`

(C)

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

(D)

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

Solution:

`:.` Foci of an ellipse `x^2/a^2 + y^2/b^2 =1` are

(ae, 0) and (-ae,0). Equation of circle
with centre ( 0,0) and radius ae is

`x^2 + y^2 = (ae)^2` [where `(ae)^2 = a^2 - b^2 ]`

`:. x^2 + y^2 = a^2 - b^2`

Correct Answer is `=>` (B) `x^2 + y^2 = a^2 -b^2`

Q 2803178048

If `A = {(x , y) : x^2 + = 25}` and `B = {(x, y) : x^2 + 9y^2 = 144 }`, then `A cap B` contains

(A)

one point

(B)

three points

(C)

two points

(D)

four points

Solution:

Clearly, A is the set of all point on

the circle `x^2 + y^2 = 25` and `B` is the set

of all points on the ellipse

`x^2 + 9y^2 = 144`. These two intersect at

four points `P, Q, R` and `S`.

Hence, `A cap B` contains four points.

Correct Answer is `=>` (D) four points

Q 2511134029

The eccentricity of the ellipse `9x^2 + 5y^2 - 30y = 0` is:
BCECE Stage 1 2015

(A)

`1/3`

(B)

`2/3`

(C)

`3/4`

(D)

None of these

Solution:

Given equation of ellipse is

`9x^2 + 5y^2 - 30y = 0`

or `9x^2 + 5 (y^2- 6y + 9) = 45`

or `x^2/5 + (y-3)^2/9 =1`

On comparing with `x^2/a^2 + y^2/b^2 =1`

Here, `a^2 =5, b^2 =9`

`:.` Eccentricity `e = sqrt (1- a^2/b^2) \ \ \ \ \ \(because b > a )`

`= sqrt(1 -5/9 ) =2/3`

Correct Answer is `=>` (B) `2/3`

Q 2520701611

An ellipse has OB as semi-minor axis, F and F' are its foci and the `angleFBF'` is a right angle. Then, the eccentricity of the ellipse is
BCECE Stage 1 2014

(A)

`1/sqrt3`

(B)

1/4

(C)

1/2

(D)

`1/sqrt2`

Solution:

Since, `angle FBF' = 90°`, then `angle OBF' = 45^o` and `angle BF'O = 45°`

`=> ae = b`

(`because Delta BOF'` is an isosceles triangle)

and `e^2 = 1 - b^2/a^2`

`=> e^2 = 1 - (a^2e^2)/a^2 => e^2 =1 -e^2`

`=> 2e^2 =1 => e = 1/sqrt2` (`because` e cannot be negative)

Correct Answer is `=>` (D) `1/sqrt2`

Q 2279180916

For the ellipse `3x^2 + 4y^2 = 5`, if `y + 3x = 0` is one diameter, the conjugate diameter
is (Note: Two diameters of an ellipse

`x^2/a^2 + y^2/b^2 =1 ` are said to be conjugate when each bisects all chords parallel to
the other).
BITSAT Mock

(A)

` 4y - x = 0`

(B)

`4y + x = 0`

(C)

`y - 4x = 0`

(D)

`y + 4x = 0`

Solution:

`x^2/(5/3) + y^2/(5/4) = 1`

`m_0 = -3 , m_1 m_0 = - b^2/a^2`

`=> m_1 = - b^2/(a^2 m_0) = - (5/4)/((5/3) (-3))`

`= 1/4`

Hence the equations of the
conjugate diameter is `y = 1/4 x`.

i.e., `4y - x = 0`

Correct Answer is `=>` (A) ` 4y - x = 0`

Q 1938112902

Find the equation of the ellipse whose centre
at `(0, 0)`, major axis on the y-axis and passes
through the points `(3, 2)` and `(1, 6)`
Class 11 Exercise 11.3 Q.No. 19

Solution:

Major axis is y-axis

Let the ellipse be `x^2/b^2 + y^2/a^2 = 1,`

since `(3, 2)` and `(1, 6)` lies on it

`:. 9/b^2 + 4/a^2 = 1` ...........(i)

` 1/b^2 + (36)/a^2 = 1` ............(ii)

Subtracting ` ( 9 - 1)/b^2 + (4 - 36)/a^2 = 0`

` => 8/b^2 - (32)/a^2 = 0 => 8a^2 = 32b^2 quad :. a^2 = 4ab^2`

Putting the value in (i), it become ` 9/b^2 + 4/(4b^2) = 1`

`:. (10)/b^2 = 1`

`:. b^2 = 10`

Now , `a^2 = 4b^2 = 4 xx 10 = 40`

`:.` Equation of the ellipse is ` x^2/(10) + y^2/(40) = 1`

Q 1729434311

If equation of the ellipse whose focus is `(1, -1)`, then directrix the line
`x - y - 3 = 0` and eccentncity `1/2` is
NCERT Exemplar

(A)

`7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0`

(B)

`7x^2 + 2xy + 7y^2 + 7 = 0`

(C)

`7x^2 + 2xy + 7y^2 + 10x - 10y - 7 = 0`

(D)

None of the above

Solution:

Given that, focus of the ellipse is `(1, - 1 )` and the equation of directrix is `x - y - 3 = 0`

and `e = 1/2`

Let `P (x, y)` and `F (1, - 1)`.

` :. (PF)/(text(Distance of P from x - y - 3 = 0 )) = 1/2`

` => sqrt((x -1)^2 + (y + 1)^2)/((| x - y - 3|)/sqrt(2)) = 1/2`

` => ( 2 [x^2 -2x + 1 + y^2 + 2y + 1]) /( x - y - 3)^2 = 1/ 4`

`=> 8x^2 - 16x + 16 + 8y^2 + 16y = x^2 + y^2 + 9 - 2xy + 6y - 6x`

` => 7x^2 +7y^2 + 2xy - 10x + 10y + 7 = 0`

Correct Answer is `=>` (A) `7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0`

Q 1535880762

The radius of the circle passing through the foci of the ellipse `\ frac {x^{2}}{16} + \ frac {y^{2}}{9} = 1` and having its centre at `(0, 3)` is
EAMCET 2015

(A)

`6`

(B)

`4`

(C)

`3`

(D)

`2`

Solution:

`(ae)^2=a^2-b^2`

`b=3, a=4`

`\therefore ae=\sqrt7`

`r = \sqrt {(ae)^{2} + b^{2}}`

`= \sqrt {7 + 9} = 4`

Correct Answer is `=>` (B) `4`

Q 1918612509

Find the coordinates of the foci, the vertices,
the length of major axis, the minor axis, the
eccentricity and the length of the latus rectum
of the ellipse `16x^2 + y^2 = 16`
Class 11 Exercise 11.3 Q.No. 8

Solution:

The ellipse is `16x^2 + y^2 = 16`

Dividing by `16` we get `x^2/1 + y^2/(16) = 1`

Major axis is along y-axis `a^2 = 16, => a = 4,`

`b^2= 1, => b = 1`

and `c^2 = a^2 - b^2 = 16 - 1 = 15 , quad :. c = sqrt(15)`

foci are `(0, ± c)` i.e. `(0, ± sqrt(15))`

vertices are `(0, ± a)` i.e. `(0, ± 4)`

Length of major axis `= 2a = 2 xx 4 = 8 ;`

Length of minor axis `= 2b = 2 xx 1 = 2`

Eccentricity `= e = c/a = sqrt(15)/4`

Length of the latus rectum ` = (2b^2)/a = (2 xx 1)/4 = 1/2`

Q 1768480305

Find the distance between the directrices of ellipse ` x^2/(36) + y^2 /(20) = 1`.
NCERT Exemplar

Solution:

The equation of ellipse is ` x^2/(36) + y^2 /(20) = 1`

On comparing this equation with ` x^2/a^2 + y^2/b^2 = 1`, we get

`a = 6, b = 2 sqrt(5)`

We know that, `b^2 = a^2 (1 - e^2)`

`=> 20 = 36 (1 - e^2)`

`=> (20)/(36) = 1 - e^2`

` :. e = sqrt(1 - (20)/(36)) = sqrt((16)/(36))`

`e= 4/6 = 2/3`

Now, directrices `= ( pm a/e , - a//e)`

`:. a/e = (6/2)/3 = ( 6 xx 3) /2 = 9`

and ` - a/e = -9`

`:.` Distance between the directrices `= | 9 - (- 9) | = 18`

Q 1778078806

If the latusrectum of an ellipse is equal to half of minor axis, then find
its eccentricity.
NCERT Exemplar

Solution:

Consider the equation of the ellipse is ` x^2/a^2 + y^2/b^2 = 1`

`:.` Length of major axis `= 2a`

Length of minor axis `= 2b`

and length of latusrectum `= (2b^2)/a`

Given that, ` (2b^2)/a = (2b)/2 `

`=> a = 2b => b = a//2`

We know that, ` b^2 = a^2 ( 1 - e^2)`

`=> (a/2)^2 = a^2 ( 1 - e^2)`

` => a^2/4 = a^2 ( 1 - e^2)`

` => 1 - e^2 = 1/4`

` => e^2 = 1 - 1/4`

`:. e = sqrt(3/4) = sqrt(3/2)`