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Tricks & Tips of Ellipse for NDA

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Finding Terms related to ellipse

Let the equation of the ellipse be `x^2/a^2+y^2/b^2=1` .....................(1)

`text(Centre : )`
In the figure, `C` is the centre of the ellipse. All chords passing through `C` are called diameter and bisected at `C`.

`text( Foci :)`
`S` and `S'` are the two foci of the ellipse and their coordinates are `( a e, 0)` and `( - a e, 0)` respecitvely.
The line containing two foci are called the `text(focal axi)`s and the distance between `S` & `S'` the focal length.

`text( Directrices :)`
`ZN` and `Z'N'` are the two direcrices of the ellipse and their equations are `x =a/e` and `x =(-a)/e` respectively. Here `Z` and `Z'` are called foot of directrix.

`text(Axes :)`
The Iine segments `A'A` and `B'B` are called the major and minor axes respectively of the ellipse.
The point of intersection of major and minor axis is called `text(centre)` of the ellipse. Major and minor axis together are called `text(principal axis of ellipse.)`

Here Semi-major axis are `CA = CA' = a`

and Semi-minor axis are `CB = CB' = b`

`text( Vertex :)`
The points where major axis meet the ellipse is called its vertices. In the given figure, `A'` and `A` are the vertices of the ellipse.

`text(Ordinate and Double Ordinates :)`
Let `P` be a point on the ellipse. From `P` we draw `PM` perpendicular to major axis of the ellipse. Produce `PM` to meet the ellipse at `P'`, then `PM` is called an ordinate and `PMP'` is called the double ordinate of the point `P`.

It is also defined as any chord perpendicular to major axis is called its double ordinate.

`text( Latus Rectum :)`
When double ordinate passes through focus then it is called the Latus rectum.

Let `L'L = 2k`, then `LS = k` so `L = (a e, k)`.

Here `L L'` and `M M'` are called latus rectum.

Since `L(a e, k)` lies on the ellipse `(1)`, therefore `(a^2e^2)/a^2+k^2/b^2 = 1` or `k^2/b^2= 1 - e^2`

or `k^2=b^2(1-e^2)=b^2 . b^2/a^2=b^2/a^2`

`:. k=b^2/a`

Hence length of semi latus rectum `LS = b^2/a=MS'`

i.e. length of the latus rectum `L L'` or `M M'=(2b^2)/a= (text(minor axis)^2)/(text(major axis))`

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

`=2e ` (distance from focus to the corresponding directrix).

And the end points of latus rectum are `L(ae,b^2/a),L'(ae,(-b^2)/a), M(-ae,b^2/a)` & `(-ae, -b^2/a)`

`text(Focal Chord :)`
A chord of the ellipse passing through its focus is called a focal chord.

`text(Focal Distance of a Point :)`
Let `P(x, y)` be any point on the ellipse

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

Then by definition of ellipse,

`SP=ePM=e(MT-PT)=e(2/e-x)=a-ex`

& `S'P=ePM'=e(M'T+PT)=e(a/e+x)=a+ex`

Hence `SP+S'P=2a`

Because of the above property, ellipse is also defined as the locus of a point which moves in a plane such that the sum of its distance from two fixed points (called foci) is a contant (Length of major axis).

This definition is called the physical definition of the ellipse.

Hence `PS + PS' = QS + QS' = TS + TS' =` length of major axis

Q 2261523425

Consider any point `P` on the ellipse `x^2/(25) + y^2/9 = 1` in the first quadrant. Let `r` and `s` represent its distances from `(4, 0)` and `(-4, 0)` respectively, then `(r + s)` is equal to
NDA Paper 1 2015
(A)

`10` units

(B)

`9` units

(C)

`8` units

(D)

`6` units

Solution:

We have, an ellipse `x^2/(25) + y^2/9 = 1`

Clearly, its foci are `( 4, 0)` and `(- 4, 0)`.

`[ ∵ foci = S (ae, 0), S' (ae, 0)]`

`:. PS + PS' = 2a =` Major axis

`=> r + s = 2(5) = 10` units
Correct Answer is `=>` (A) `10` units
Q 1782145937

What is the sum of the major and minor axes of
the ellipse whose eccentricity is `4//5` and length of
latusrectum is `14.4` units?
NDA Paper 1 2014
(A)

32 units

(B)

48 units

(C)

64 units

(D)

None of these

Solution:

We know that, length of major axes of an

ellipse `= 2a`

and length of minor axes of an ellipse `= 2b`

Given that,

eccentricity of an ellipse `= 4 //5 = e` .........(1)

and length of latusrectum of an ellipse `= 14.4` units

`=> ( 2b^2)/a = 14.4`

` => (b^2)/a = 7.2`

` => b^2 = 7.2 a` .......(2)

Since, eccentricity of an ellipse,

` b^2 = a^2 (1-e^ 2)`

` => 7.2a = a^2 [1 - (4/5)^2 ] ` [from Eqs. (i) and (ii)]

` => 7.2a = a^2 (1- (16)/(25))`

` => 7.2 a = a^2 xx 9/(25)`

` => 9a^2 - 7.2 xx 25a = 0`

` => 9a^2 -36 xx 5a = 0`

` => 9a (a-20) = 0`

` => a = 20`

Put the value of a in Eq. (2), we get

` b^2 =7.2 xx 20`

`=> b^2 = 72 xx 2 = 144`

` => b^2 = (12)^2`

`:. b = 12`

Hence, the sum of the major and minor axes

` = 2a + 2b`

`= 2 (a + b) = 2 (20 + 12)`

` = 2 xx 32`

` = 64` units
Correct Answer is `=>` (C) 64 units
Q 1629167911

What is the length of the latus rectum of an ellipse
`25x^2 + 16y^2 = 400?`


NDA Paper 1 2014
(A)

` (25)/2`

(B)

` (24)/4`

(C)

` (16)/5`

(D)

` (32)/5`

Solution:

Equation of ellipse is `25x^2 + 16y^2 = 400`.

`=> x^2/(16) + y^2/(25) = 1`

Here, `a^2 = 16` and `b^2 = 25`

:. Length of latus rectum `= (2a^2)/b = (2 xx 16)/5 = (32)/5`
Correct Answer is `=>` (D) ` (32)/5`
Q 2328078801

The equation of the ellipse whose vertices are at

`(pm 5,0)` and foci at `(pm 4,0)` is
NDA Paper 1 2013
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

Given that,

Foci of an ellipse `=( pm 4, 0) = (pm ae, 0)`

`=> ae=4` ...................(i)

and vertices of an ellipse `=( pm 5, 0)= (pm a, 0)`

`=> a=5` .....................(ii)

From Eqs. (i) and (ii),

`e= 4/5` ......................(iii)

Now, we have a relation

`b^2 = a^2 (1-e^2) => b^2 =25 (1- 16/25)`

`=> b^2 = 25 * 9/25 =>b^2 = 9`

`:. b= pm 3`

`:.` Required equation of an ellipse,

`x^2/a^2 + y^2/b^2 =1`

`:. x^2/25 +y^2/9 =1`
Correct Answer is `=>` (A) `x^2/25 + y^2/9 =1`
Q 2308078808

The sum of the focal distances of a point on the

ellipse `x^2/4+y^2/9 =1` is
NDA Paper 1 2012
(A)

`4` units

(B)

`6` units

(C)

`8` units

(D)

`10` units

Solution:

Since. the sum of focal distances of a point on the

ellipse `x^2/a^2 + y^2/b^2 =1` is equal to `2b`. when `b >a`.

`:. a^2=4` and `b^2 = 9 => a =2` and `b=3`

`:.` Sum of the focal distances `= 2 xx 3 = 6` units
Correct Answer is `=>` (B) `6` units
Q 2348280103

What is the sum of the focal distances of a point

of an ellipse `x^2/a^2 + y^2/b^2 =1?`
NDA Paper 1 2011
(A)

`a`

(B)

`b`

(C)

`2a`

(D)

`2b`

Solution:

The sum of the focal distances of a point of an ellipse

`x^2/a^2 + y^2/b^2 =1` is `2a` (by property)
Correct Answer is `=>` (C) `2a`
Q 2358280104

If the latusrectum of an ellipse is equal to half its
minor axis, then what is its eccentricity?
NDA Paper 1 2010
(A)

`1/2`

(B)

`sqrt (3)`

(C)

`sqrt(3)/2`

(D)

`1/sqrt(2)`

Solution:

Now, the length of latusrectum of an ellipse `= (2b^2)/a` and

its length of minor axis ` =2b`

According to the question,

`(2b^2)/a =1/2 (2b) => (2b^2)/a = b`

` => 2b = a => 4b^2 = a^2` `( :. b ne 0 )`

`=> 4a^2 (1-e^2) = a^2` `[ :. a^2 = b^2 (1-e^2) ]`

`=> e^2 = 3/4`

`:. e= sqrt (3)/2`
Correct Answer is `=>` (B) `sqrt (3)`
Q 2388280107

If `(4, 0)` and `(- 4, 0)` are the foci of an ellipse and the
semi-minor axis is `3`, then the ellipse passes
through which one of the following points?
NDA Paper 1 2010
(A)

`(2,0)`

(B)

`(0,5)`

(C)

`(0,0)`

(D)

`(5,0)`

Solution:

Since, the foci of an ellipse are `( 4, 0)` and `( -4, 0)`.

`:. 2 ae = 8 => ae =4`

and `b =3` (semi-minor axis)

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

`=> (4/a)^2 = (1- 9/a^2)`

`=> 16/a^2 = (a^2 - 9)/a^2 => a^2 = 25`

`:. a =5`

Thus, the equation of an ellipse is

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

Which is satisfied by `(5, 0)`.
Hence, the ellipse passes through `(5, 0)`.
Correct Answer is `=>` (D) `(5,0)`
Q 2378380206

What is the eccentricity of an ellipse, if its
latusrectum is equal to one-half of its minor axis?
NDA Paper 1 2009
(A)

`1/4`

(B)

`1/2`

(C)

`sqrt(3)/4`

(D)

`sqrt(3)/2`

Solution:

Length of the latusrectum of an ellipse `= (2b^2)/a`

and length of the minor axis `= 2b`

`:. 1/2 (2b ) = (2b^2)/a => a =2b`

`( :. b ne 0` according to the question)

Also, `e =sqrt (1-b^2/a^2)`

`= sqrt (1- b^2/(4b^2)) = sqrt (3/4) = (sqrt (3))/2`
Correct Answer is `=>` (D) `sqrt(3)/2`
Q 2308380208

What is the sum of focal radii of any point on an
ellipse equal to?
NDA Paper 1 2009
(A)

Length of latusrectum

(B)

Length of major axis

(C)

Length of minor axis

(D)

Length of semi-latusrectum

Solution:

We know that, the sum of focal radii of any point on an
ellipse is equal to length of major axis. i.e., sum of focal radii

`=(a+ x) +(a-x)`

`= 2a =` Major axis
Correct Answer is `=>` (B) Length of major axis
Q 1988612507

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.

` x^2/(100) + y^2/(400) = 1`
Class 11 Exercise 11.3 Q.No. 6
Solution:

` x^2/(100) + y^2/(400) = 1` is the equation of ellipse.

Major axis is along y-axis

`a^2 = 400, quad :. a = 20, b^2 = 100 quad :. b = 10`

`c^2 = a^2 - b^2 = 400 - 100 = 300 quad :. c = 10 sqrt(3)`

Vertices are `(0, ± a)` i.e., `(0, ± 20)`

`:.` Foci are `(0, ± c)` i.e., `(0, ± 10 sqrt(3))`

Length of major axis `= 2a = 2 xx 20 = 40`

Length of minor axis `= 2b = 2 xx 10 = 20`

Eccentricity, `e = c/a = (10sqrt(3))/(20) = sqrt(3)/2`

Length of Latus rectum `= (2b^2)/a = ( 2 xx 100)/(20) = 10`
Q 1978612506

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.

`x^2/(49) + y^2/(36) = 1`
Class 11 Exercise 11.3 Q.No. 5
Solution:

`x^2/(49) + y^2/(36) = 1` is the equation of ellipse.

`a^2 = 49,` major axis is along x-axis

`a^2 = 49 quad :. a = 7, b^2 = 36, b = 6`

`:. c^2 = a^2 - b^2 = 49 - 36 = 13 quad :. c = sqrt(13)`

Foci are `(pm c, 0)` i.e., `(± sqrt(13), 0)`

Vertices are `(± a, 0)` i.e., `(pm 7, 0)`

Length of major axis `= 2b = 2 xx 7 = 14`

Length of minor axis `= 2b = 2 xx 6 = 12`

Length of Latus rectum `= (2b^2)/a = (2 xx 36)/7 = (72)/7`

Ecentricity, `e = c/a = sqrt(13)/7`

Different standard forms of ellipse

Q 1782145937

What is the sum of the major and minor axes of
the ellipse whose eccentricity is `4//5` and length of
latusrectum is `14.4` units?
NDA Paper 1 2014
(A)

32 units

(B)

48 units

(C)

64 units

(D)

None of these

Solution:

We know that, length of major axes of an

ellipse `= 2a`

and length of minor axes of an ellipse `= 2b`

Given that,

eccentricity of an ellipse `= 4 //5 = e` .........(1)

and length of latusrectum of an ellipse `= 14.4` units

`=> ( 2b^2)/a = 14.4`

` => (b^2)/a = 7.2`

` => b^2 = 7.2 a` .......(2)

Since, eccentricity of an ellipse,

` b^2 = a^2 (1-e^ 2)`

` => 7.2a = a^2 [1 - (4/5)^2 ] ` [from Eqs. (i) and (ii)]

` => 7.2a = a^2 (1- (16)/(25))`

` => 7.2 a = a^2 xx 9/(25)`

` => 9a^2 - 7.2 xx 25a = 0`

` => 9a^2 -36 xx 5a = 0`

` => 9a (a-20) = 0`

` => a = 20`

Put the value of a in Eq. (2), we get

` b^2 =7.2 xx 20`

`=> b^2 = 72 xx 2 = 144`

` => b^2 = (12)^2`

`:. b = 12`

Hence, the sum of the major and minor axes

` = 2a + 2b`

`= 2 (a + b) = 2 (20 + 12)`

` = 2 xx 32`

` = 64` units
Correct Answer is `=>` (C) 64 units
Q 1753223144

Consider an ellipse, `x^2/a^2 + y^2/b^2 = 1`

What is the area included between the ellipse and the
greatest rectangle inscribed in the ellipse?
NDA Paper 1 2014
(A)

`ab (pi - 1)`

(B)

`2ab (pi - 1)`

(C)

`ab (pi - 2)`

(D)

None of these

Solution:

Given ellipse, `x^2/a^2 + y^2/b^2 = 1`

We know that,

Area of the ellipse `x^2/a^2 + y^2/b^2 = 1` is, `Delta ' = pi ab`

`:.` Required area =Area of shaded region

=Area of an ellipse

-Area of greatest rectangle

`= pi ab - 2ab`

` = ab (pi -2)`
Correct Answer is `=>` (C) `ab (pi - 2)`
Q 2318078809

The eccentricity e of an ellipse satisfies the
condition
NDA Paper 1 2012
(A)

`e < 0`

(B)

`0 < e < 1`

(C)

`e=1`

(D)

`e > 1`

Solution:

The eccentricity of ellipse lies between `0` and `1` .
Correct Answer is `=>` (B) `0 < e < 1`
Q 2318178900

What is the eccentricity of the conic

`4x^2 +9y^2 = 144`?

NDA Paper 1 2012
(A)

`sqrt (5)/3`

(B)

`sqrt (5)/4`

(C)

`3/sqrt(5)`

(D)

`2/3`

Solution:

The given equation of conic is

`4x^2 +9y^2 =144`

`=> x^2/36 +y^2/16 =1`

which represent an ellipse. here `a> b`.

`:. a^2 = 36 => a =6`

`b^2 =16 => b =4`

Now, eccentricity, `b^2 = a^2 (1 - e^2 )`

`=>16 = 36 (1-e^2)`

`=> 4/9 =1-e^2 =>e^2 = 1 -4/9 =5/9`

`:. e= sqrt(5)/3`
Correct Answer is `=>` (A) `sqrt (5)/3`
Q 2328480301

Consider the ellipse `x^2/a^2 +y^2/b^2 =1 (b > a)`. Then,

which one of the following is correct?
NDA Paper 1 2008
(A)

Real foci do not exist

(B)

Foci are `(pm ae , 0)`

(C)

Foci are `(pm be, 0)`

(D)

Foci are `(0, pm be)`

Solution:

Given, equation of ellipse

`x^2/a^2 +y^2/b^2 =1`

Since , `b > a`

`:.` Foci `= (0, pm be)` (by property)
Correct Answer is `=>` (D) Foci are `(0, pm be)`
Q 2308480308

Which one of the following is correct?

The eccentricity of the conic `x^2/(a^2 + lambda) +y^2/(b^2 +lambda ) =1`

(where, `lambda ge 0`)
NDA Paper 1 2008
(A)

increases with increase in `lambda`

(B)

decreases with increase in `lambda`

(C)

does not change with `lambda`

(D)

None of the above

Solution:

The given equation of the conic is

`x^2/(a^2 + lambda) +y^2/(b^2 +lambda) = 1 , (lambda ge 0)`

Here, `A^2 = a^2 +lambda` and `B^2 = a^2 + lambda`

Eccentricity , `e= sqrt (1- B^2/A^2) = sqrt (1- (b^2 +lambda)/(a^2 +lambda))`

`= sqrt ((a^2 + lambda -b^2 - lambda^2 )/(a^2 + lambda)) = sqrt ((a^2 - b^2)/(a^2 + lambda))`

From above it is clear that, when `lambda`. increases, the eccentricity
decreases.
Correct Answer is `=>` (B) decreases with increase in `lambda`
Q 2339278112

If the ellipse `x^2/169 +y^2/25 =1` has the same

eccentricity as the ellipse `x^2/a^2 + y^2/b^2 =1`, then what

is the ratio of `a` to `b`?
NDA Paper 1 2009
(A)

`5/13`

(B)

`13/5`

(C)

`7/8`

(D)

`8/7`

Solution:

Given ellipse is `x^2/169 + y^2/25 =1`

`:. e= sqrt (1-25/169) = 12/13 ` `( :. e=sqrt (1- b^2/a^2) )`

Also, ellipse is `x^2/a^2 + y^2/b^2 =1`

`:. e = sqrt (1-b^2/a^2)` (according to the question)

`:. 12/13 = sqrt (1-b^2/a^2)`

`=> b^2/a^2 =1 -144/169 = 25/169`

`=> a/b = 13/5`
Correct Answer is `=>` (B) `13/5`

Finding equation of ellipse

See examples
Q 2773180946

What is the equation of the ellipse having foci `(±2, 0)` and the eccentricity `1/4 ?`
NDA Paper 1 2017
(A)

`x^2/64 +y^2/60 = 1`

(B)

`x^2/60+y^2/64 = 1`

(C)

`x^2/20+y^2/24 =1`

(D)

`x^2/24+y^2/20 = 1`

Solution:

foci `(pm ae , 0)`

`ae= 2 , e=1/4`

`a=8`

`b^2 =64 (1- 1/16)=60`

`x^2/a^2 +y^2/b^2 =1`

`x^2/64 +y^2/60 =1`
Correct Answer is `=>` (A) `x^2/64 +y^2/60 = 1`
Q 2318380200

What are the equation of the directrices of the
ellipse `25x^2 + 16 y^2 = 400`?
NDA Paper 1 2010
(A)

`3x pm 25 =0`

(B)

`3y pm 25 =0`

(C)

`x pm 15 =0`

(D)

`y pm 25 =0`

Solution:

The equation of ellipse can be rewritten as

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

Here, `a=4` and `b =5` `( b > a )`

`:.` Equation of the directrices are

` y = pm b/e = pm 5/(sqrt (1- 16/25) ) = pm25/3` `[ :. a^2 = b^2 (1-e^2) ]`

`=> 3y pm 25 =0`
Correct Answer is `=>` (B) `3y pm 25 =0`
Q 2761780625

What is the equation of the hyperbola having latus rectum and eccentricity 8 and `3/sqrt5` respectively ?
NDA Paper 1 2016
(A)

`x^2/25 - y^2/20 =1`

(B)

`x^2/40 - y^2/20 =1`

(C)

`x^2/40 - y^2/30 =1`

(D)

`x^2/30 - y^2/25 =1`

Solution:

`(2b^2)/a =8`

`b^2=4a`

and `e= 3/(sqrt 5)`

For Hyperbola

`b^2 = a^2 [e^2-1]`

`4a^2 = a^2 [(3/(sqrt 5))^2 -1]`

`a=5 , b= sqrt 20`

So, equation of hyperbola

`(x^2)/25 -(y^2)/20 =1`
Correct Answer is `=>` (A) `x^2/25 - y^2/20 =1`
Q 1918712609

Find the equation for the ellipse that satisfies the given conditions:
Length of minor axis `16`, foci `(0, + 6)`
Class 11 Exercise 11.3 Q.No. 16
Solution:

Since the foci are on the y-axis, the major axis

will be on y-axis. The equation will be of the form

`x^2/b^2 + y^2/a^2 = 1`

Given `2b = 16` or `b = 8` and `c = 16`

We know that `c^2 = a^2 - b^2` or `36 = a^2 - 64`

or `a^2 = 100` or `a = 10`

Hence, the equation of the ellipse is

`x^2/(64) + y^2/(100) = 1`
Q 1908712608

Find the equation for the ellipse that satisfies
the given conditions:
Length of major axis `26,` Foci `(± 5, 0)`
Class 11 Exercise 11.3 Q.No. 15
Solution:

Length of major axis `= 2a = 26 quad :. a = 13`

Foci are `(± 5, 0), c = 5,`

`:. b^2 = a^2 - c^2`

`= 169 - 25 = 144`

Major axis is `x`-axis.

Equation of ellipse is `x^2/(169) + y^2/(144) = 1`
Q 1988712607

Find the equation for the ellipse that satisfies
the given conditions :
Ends of major axis `(0, ± sqrt(5) )`, ends of minor axis
`(± 1, 0)`.
Class 11 Exercise 11.3 Q.No. 14
Solution:

Ends of major axis `(0, ± sqrt(5) )`.

Major axis is they-axis and `a = sqrt(5)`

Ends of minor axis are `(± 1, 0)`

`b = 1`

Equation of ellipse is `x^2/1 + y^2/5 = 1`
Q 2368280105

The ellipse `x^2/169 + y^2/25 =1` has the same eccentricity

as the ellipse `x^2/a^2 + y^2/b^2 =1` what is the ratio of `a` to `b` ?
NDA Paper 1 2010
(A)

`5/13`

(B)

`13/5`

(C)

`7/8`

(D)

`8/7`

Solution:

Given ellipse is `x^2/169 + y^2/25 =1`

`e= sqrt (1- 25/169) =12/13` `( :. e= sqrt (1- b^2/a^2) )`

Also, be the eccentricity of the ellipse `x^2/a^2 + y^2/b^2 =1`.

According to the question,

`e =sqrt (1-b^2/a^2) => 12/13 = sqrt (1-b^2/a^2)`

`=>b^2/a^2 = 1- 144/169 =25/169`

`:. a/b = 13/5`
Correct Answer is `=>` (B) `13/5`

Position of a Point W.r.t. an Ellipse :

Let `S(x, y) = x^2/a^2+y^2/b^2-1` be the given ellipse and `P(x_1, y_1)` is the given point.

(i) If `S(x_1, y_1) > 0` then `P(x_1,y_1)` lie outside the ellipse.

(ii) If `S(x_1,y_1) < 0` then `P(x_1, y_1)` lie inside the ellipse.

(iii) If `S(x_1, y_1) = 0` then `P(x_1,y_1)` lie on the ellipse.

This result holds true for circle and parabola also.

Q 2348380203

Let `E` be the ellipse `x^2/9 + y^2/4 =1` and `C` be the circle

`x^2 + y^2 =9`. Let `P= (1,2)` and `Q= (2,1)`, then which
one of the following is correct?
NDA Paper 1 2010
(A)

Q lies inside C but outside E

(B)

Q lies outside both C and E

(C)

P lies inside both C and E

(D)

P lies inside C but outside E

Solution:

For a point `P(1, 2)`

`E = 4(1)^2 + 9(2)^2 - 36`

`= 40 -36 > 0`

and `C = 1^2 + 2^2 -9`

`= 5 -9 < 0`

So, the point `P` lies outside of `E` and inside of `C`.
For a point `Q(2,1)`,

`E = 4 (2)^2 + 9(1)^2 -36`

`= 16 + 9 -36 < 0`

and `(2)^2 + (1)^2 -9 = 4 +1 - 9 < 0`

Hence, the point `Q` lies inside of `E` and `C`.
Correct Answer is `=>` (D) P lies inside C but outside E

Miscellaneous

Q 2751080824

If the ellipse `9x^2+16y^2 = 144` intercepts the line `3x+4y = 12` then what is the length of the chord so formed ?
NDA Paper 1 2016
(A)

`5` units

(B)

`6` units

(C)

`8` units

(D)

`10` units

Solution:

`x^2/16+y^2/9 = 1` , Line `x/4+y/3 = 1`

Line intersect to given ellipse

`(4 , 0 ) & ( 0 , 3)`

Length of chord `AB = sqrt((4-0)^2+(3-0)^2)`

`AB = 5`
Correct Answer is `=>` (A) `5` units
Q 1783123047

Consider an ellipse, `x^2/a^2 + y^2/b^2 = 1`

What is the area of the greatest rectangle that can be
inscribed in the ellipse?
NDA Paper 1 2014
(A)

`ab`

(B)

`2ab`

(C)

`(ab)/2`

(D)

`sqrt(ab)`

Solution:

Given ellipse, `x^2/a^2 + y^2/b^2 = 1` .........(1)

Let` A (a cos theta , b sin theta)` be any point on an ellipse.

(1st quadrant)

Coordinate of `B`

`= {a cos (pi - theta ), b sin ( pi - theta )}`

`= (-a cos theta , b sin theta )` (2nd quadrant)

and coordinate of `C`

`= {a cos (pi + theta), b sin (pi + theta)`

`= (- a cos theta , - b sin theta )` (3 rd quadrant)

and coordinate of `D`

` = {a cos (2pi - theta), b sin (2pi - theta)}`

`= (a cos theta ,- b sin theta)` (4 quadrant)

Now, area of rectangle `ABCD,`

`Delta = CD xx AD`

`= (a cos theta + a cos theta ) xx (b sin theta + b sin theta )`

` = 2 a cos theta xx 2 b sin theta`

` = 2ab sin 2theta`

Here, area of rectangle will be greatest, when `sin 2theta` have

its maximum value i.e., `sin 2 theta = 1`.

`:. Delta = 2ab xx 1 = 2ab`

Hence, area of greatest rectangle is equal to `2ab`, when

`sin 2theta = 1`.
Correct Answer is `=>` (B) `2ab`
Q 2308280108

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?
NDA Paper 1 2010
(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:

Since, the foci of an ellipse `x^2/a^2 +y^2/b^2 =1` are `(ae , 0 )` and

`(-ae , 0)`

`:.` Equation of circle whose end points of diameter are `(ae ,0 )`

and `(-ae , 0)` is

`(x - ae )(x + ae) + (y - 0)(y - 0) = 0`

`=> x^2 - a^2 e^2 + y^2 =0`

`=> x^2 + y^2 -a^2 (1- b^2/a^2) =0`

`[ :. e^2 = (1-b^2/a^2) ]`

`=> x^2 +y^2 -a^2 + b^2 =0`

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