# Class 11 RD Sharma Solutions – Chapter 26 Ellipse – Exercise 26.1 | Set 1

Last Updated : 01 Feb, 2023

### Question 1. Find the equation of the ellipse whose focus is (1,â€“2) and directrix is 3x â€“ 2y + 5 = 0, and eccentricity is 1/2.

Solution:

Given that,

Focus is (1, -2)

directrix is 3x â€“ 2y + 5 = 0,

eccentricity(e) is 1/2.

As we know, SP = ePM

â‡’ SP = PM/2

â‡’ (SP)2 = 1/4(PM)2

â‡’ 4(SP)2 = (PM)2

â‡’ 4[(x – 1)2 + (y + 2)2] =

â‡’ 4[x2 + 1 – 2x + y2 + 4 + 4y] =

â‡’ 52[x2 + 1 – 2x + y2 + 4 + 4y] = (3x – 2y + 5)2

â‡’ 52[x2 + 1 – 2x + y2 + 4 + 4y] = 9x2 + 4y2 + 25 – 12xy – 20y + 30x

Thus 43x2 + 43y2 + 12xy â€“ 134x + 228y + 235 = 0 is the required equation.

### (i) Focus is (0, 1), directrix is x + y = 0 and e = 1/2.

Solution:

Given that,

focus is (0, 1),

directrix is x + y = 0

eccentricity(e) is 1/2

As we know, SP = ePM

â‡’ SP = PM/2

â‡’ (SP)2 = 1/4(PM)2

â‡’ 4(SP)2 = (PM)2

â‡’ 4[(x – 0)2 + (y – 1)2] =

â‡’ 8[x2 + y2 – 2y + 1] = x2 + y2 + 2xy

Thus 7x2 + 7y2 â€“ 2xy â€“ 16y + 8 = 0 is the required equation.

### (ii) Focus is (â€“1,1), directrix is x â€“ y + 3 = 0 and e = 1/2.

Solution:

Given that,

focus is (-1, 1),

directrix is x – y + 3 = 0

eccentricity(e) is 1/2

As we know, SP = ePM

â‡’ SP = PM/2

â‡’ (SP)2 = 1/4(PM)2

â‡’ 4(SP)2 = (PM)2

â‡’ 4[(x + 1)2 + (y – 1)2] =

â‡’ 4[(x + 1)2 + (y – 1)2] = (x – y + 3)2/2

â‡’ 8[(x + 1)2 + (y – 1)2] = (x – y + 3)2

â‡’ 8[(x + 1)2 + (y – 1)2] = x2 + y2 + 9 – 6y – 2xy + 6x

â‡’ 8[(x2 + 1 + 2x) + (y2 + 1 – 2y)] = x2 + y2 + 9 – 6y – 2xy + 6x

â‡’ 8[x2 + y2 +  2 + 2x – 2y] = x2 + y2 + 9 – 6y – 2xy + 6x

Thus 7x2 + 7y2 + 2xy + 10x â€“ 10y + 7 = 0 is the required equation.

### (iii) Focus is (â€“2,3), directrix is 2x + 3y + 4 = 0 and e = 4/5.

Solution:

Given that,

focus is (-2, 3),

directrix is 2x + 3y + 4 = 0

eccentricity(e) is 4/5

As we know, SP = ePM

â‡’ SP = 4/5(PM)

â‡’ (SP)2 = 16/25(PM)2

â‡’ 25(SP)2 = 16(PM)2

â‡’ 25[(x + 2)2 + (y – 3)2] =

â‡’ 25[(x + 2)2 + (y – 3)2] = 16(2x + 3y + 4)2/13

Thus 325[x2 + y2 + 4x – 6y + 13] = 16(2x + 3y + 4)2 is the required equation.

### (iv) Focus is (1, 2), directrix is 3x + 4y â€“ 5 = 0 and e = 1/2.

Solution:

Given that,

focus is (1, 2),

directrix is 3x + 4y â€“ 5 = 0

eccentricity(e) is 1/2

We know, SP = ePM

â‡’ SP = PM/2

â‡’ (SP)2 = 1/4(PM)2

â‡’ 4(SP)2 = (PM)2

â‡’ 4[(x – 1)2 + (y – 2)2] =

â‡’ 4[(x – 1)2 + (y – 2)2] = (3x + 4y – 5)2/25

â‡’ 100[(x – 1)2 + (y – 2)2] = 9x2 + 16y2 + 25 + 24xy – 40y – 30x

â‡’ 100[(x2 + 1 – 2x) + (y2 + 4 – 4y)] = 9x2 + 16y2 + 25 + 24xy – 40y – 30x

Thus 91x2 + 84y2 â€“ 24xy â€“ 170x â€“ 360y + 475 = 0 is the required equation.

### (i) 4x2 + 9y2 = 1

Solution:

Given that 4x2 + 9y2 = 1

So,

â‡’ Eccentricity =

= âˆš5/3

Length of latus rectum =

= 4/9

Foci are(âˆš5/6; 0) and (-âˆš5/6; 0)

### (ii) 5x2 + 4y2 = 1

Solution:

Given that 5x2 + 4y2 = 1

So,

â‡’ Eccentricity =

= 1/âˆš5

Length of latus rectum =

= 4/5

Foci are (0; 1/2âˆš5) and (0; -1/2âˆš5)

### (iii) 4x2 + 3y2 = 1

Solution:

Given that 4x2 + 3y2 = 1

So,

â‡’ Eccentricity =

= 1/2

Length of latus rectum = 2a2/b

= âˆš3/2

Foci are (0; 1/2âˆš3)  and (0; -1/2âˆš3).

### (iv) 25x2 + 16y2 = 1600

Solution:

Given that 25x2 + 16y2 = 1600

â‡’

So,

â‡’ Eccentricity =

= 3/5

â‡’ Coordinates of foci are (0, 6) and (0, â€“6).

â‡’ Length of latus rectum = 2a2/b

= 2 x (64/10)

= 64/5

### (v) 9x2 + 25y2 = 225

Solution:

Given that 9x2 + 25y2 = 225

=> \frac{9x^2}{225}+\frac{25y^2}{225}=1

=>

Clearly, a = 5 and b = 3.

So,

â‡’ Eccentricity =

= 4/5

â‡’ Coordinates of foci are (4, 0) and (â€“4, 0).

â‡’ Length of latus rectum = 2b2/a

= 2 x (9/5)

= 18/5

### Question 4. Find the equation of the ellipse passing through the point (â€“3, 1) and has eccentricity .

Solution:

Let the equation of the plane be:

…(i)

It is given that the ellipse pass through the point (â€“3, 1), so,

…(ii)

â‡’

â‡’

â‡’ b2/a2 = 3/5

â‡’ b2 = 3a2/5

â‡’ b2 = 3a2/5 ……(iii)

Now put the value of b2 in equation (ii), we get

9 + 5/3 = a2

a2 = 32/3

Now put the value of a2 in eq(iii), we get,

b2 = 3/5 x 32/3 = 32/5

Now put the a2 and b2 in eq(i), we get,

Thus 3x2 + 5y2 = 32 is the required equation of the plane.

### (i) e = 1/2 and foci (Â±2, 0).

Solution:

Let the equation of the ellipse be:

…(i)

Now, ae = 2

or, a2 = 16

Now, b2 = a2(1 â€“ e2)

â‡’ b2 = 16(1 â€“ (1/2)2)

â‡’ b2 = 12

Now put the a2 and b2 in eq(i), we get,

Thus 3x2 + 4y2 = 48 is the equation of the ellipse.

### (ii) e = 2/3 and length of latus- rectum = 5

Solution:

Let the equation of the ellipse be:

…(i)

Now,

â‡’ 2b2/a = 5

â‡’ b2 = 5a/2  ….(ii)

Since, b2 = a2(1 – e2)

â‡’ 5a/2  = a2(1 – (2/3)2)

â‡’ a = 9/2

â‡’ a2 = 81/4

Now put the value of a in eq(ii), we get

b2 = 5/2 x 9/2

b2 = 45/4

Now put the a2 and b2 in eq(i), we get,

Thus 20x2 + 36y2 = 405 is the required equation.

### (iii) e = 1/2 and semi-major axis = 4

Solution:

Let the equation of the ellipse be:

….(i)

Semi – major axis = a = 4

â‡’ a2 = 16

We know, a2 = b2(1 – e2)

â‡’ 16 = b2(1 – (12/22))

â‡’ b2 = 12

Now put the a2 and b2 in eq(i), we get,

â‡’

Thus 3x2 + 4y2 = 48 is the required equation.

### (iv) e = 1/2 and major axis = 12

Solution:

Let the equation of the ellipse be:

….(i)

Given, 2a = 12

â‡’ a = 6

We know,

â‡’

â‡’ b2 = 27

On substituting the values of a2 and b2 in eq(i), we get,

â‡’

â‡’

Thus 3x2 + 4y2 = 108 is the equation of the ellipse.

### (v) It passes through (1, 4) and (â€“6, 1).

Solution:

Let the equation of the ellipse be:

….(i)

It is given that ellipse passes through (1, 4) and (â€“6, 1), we get

Let p = 1/a2 and r = 1/b2

â‡’ p + 16r = 1       ……(ii)

Since the ellipse also passes through the point (â€“6, 1), we have

â‡’ 36p + r = 1      ……(iii)

On solving eq(ii) and (iii), we have:

p = 3/115; r = 7/115

On substituting the values in eq(i), we get;

â‡’

Thus 3x2 + 7y2 = 115 is the required equation.

### (vi) Its vertices are (Â±5, 0) and foci(Â±4, 0)

Solution:

Let the equation of the ellipse be:

….(i)

Given that a = 5 and ae = 4

Thus, e = 4/5

Now, b2 = a2(1 – e2)

â‡’ b2 = 25(1 – 16/25)

â‡’ b2 = 9

On substituting the values of a2 and b2 in eq(i), we get,

Thus  is the required equation.

### (vii) Vertices are (0,  Â±13) and foci(0,  Â±5)

Solution:

Let the equation of the ellipse be:

….(i)

Given: b = 13 and be = 5

Hence, e = 5/13

Now, a2 = b2(1 – e2)

â‡’ a2 = 169(1 – 25/169)

â‡’ a2 = 144

On substituting the values of a2 and b2 in eq(i), we get,

Thus  is the required equation.

### (viii) Its vertices are (Â±6, 0) and foci(Â±4, 0)

Solution:

Let the equation of the ellipse be:

….(i)

Given: a = 6 and ae = 4

Thus, e = 2/3

Now, b2 = a2(1 – e2)

â‡’ b2 = 36(1 – 16/36)

â‡’ b2 = 20

On substituting the values of a2 and b2 in eq(i), we get,

Thus  is the required equation.

### (ix) Ends of the major axis (Â±3, 0) and ends of the minor axis (0, Â±2).

Solution:

Let the equation of the ellipse be …..(i)

Ends of major axis = (Â±3, 0)

Ends of minor axis = (0, Â±2)

Since the ends of the major and minor axes are (Â±a, 0) and (0, Â±b) respectively.

Hence, a = 3 and b = 2

So, a2 = 9, b2 = 4

On substituting the values of a2 and b2 in eq(i), we get,

Thus, is the required equation.

### (x) Ends of the major axis (0, Â±âˆš5) and ends of the minor axis (Â±1, 0).

Solution:

Let the equation of the ellipse be  ….(i)

Ends of major axis = (0, Â±âˆš5)

Ends of minor axis = (Â±1, 0)

Since the ends of the major and minor axes are (Â±a, 0) and (0, Â±b) respectively.

Hence, a = 1 and b = âˆš5

So, a2 = 1, b2 = 5

On substituting the values of a2 and b2 in eq(i), we get,

Thus, is the required equation.

### (xi) Length of the major axis is 26 and foci (Â±5, 0).

Solution:

Let the equation of the ellipse be:

….(i)

Given that the length of major axis = 26 and foci (Â±5, 0).

We have, 2a = 26

â‡’ a = 13

â‡’ a2 = 169

Also, ae = 5

â‡’ e = 5/13

We know,

â‡’

â‡’ b2 = 144

On substituting the values of a2 and b2 in eq(i), we get,

Thus  is the required equation.

### (xii) Length of minor axis is 16 and foci(0, Â±6)

Solution:

Let the equation of the ellipse be:

….(i)

Given that, length of minor axis is 16

2a = 16

a = 8

a2 = 64

Now the coordinates of foci are (0, Â±be)

So, be = 6

(be)2 = 36

We know, a2 = b2(1 – e2)

a2 = b2 – b2e2

64 = b2 – 36

b2 = 100

On substituting the values of a2 and b2 in eq(i), we get,

Thus,  is the required equation.

### (xiii) Foci are (Â±3, 0) and a = 4

Solution:

Let the equation of the ellipse be:

….(i)

Given that, ae = 3 and a = 4

Thus, e = 3/4 and a2 = 16

We know,

â‡’

â‡’ b2 = 7

On substituting the values of a2 and b2 in eq(i), we get,

Thus  is the required equation of the ellipse.

### Question 6. Find the equation of the ellipse whose foci are (Â±4, 0), e = 1/3.

Solution:

Let the equation of the ellipse be   ….(i)

Then the coordinates of the foci are (Â±a, 0).

We have, ae = 4 and  e = 1/3

Thus, a = 12

and a2 = 144

We know that, b2 = a2(1 – e2)

â‡’ b2 = 144(1 – (1/3)2)

â‡’ b2 = 144(8/9)

â‡’ b2 = 128

On substituting the values of a2 and b2 in eq(i), we get,

Thus,  is the required equation.

### Question 7. Find the equation of the ellipse in the standard form whose minor axis is equal to the distance between the foci and whose latus rectum is 10.

Solution:

Given that, the coordinates of foci are (Â±ae, 0).

2b = 2ae

â‡’ b = ae

â‡’ b2 = a2e2 ….(i)

Also given that the latus rectum is 10

So, 2b2/a = 10

b2= 5a  ….(ii)

As we know that b2 = a2(1 – e2

â‡’ b2 = a2(1 – e2

â‡’ b2 = a2 – a2e2

â‡’ b2 = a2 – b2

â‡’ 2b2 = a2

â‡’ b2 = a2/2  ….(iii)

Now put the value of b2 in eq(ii), we get

a2/2 = 5a

a = 10

So, a2 = 100

Now put the value a2 of in eq(iii), we get

b2 = 100/2

b2 = 50

On substituting the values of a2 and b2 in eq(i), we get,

Hence, x2 + 2y2 = 100 is the required equation.

### Question 8. Find the equation of the ellipse whose centre is (-2, 3) and whose semi- axis are 3 and 2 when the major axis is (i) parallel to the x-axis (ii) parallel to the y-axis.

Solution:

(i) When the major axis is parallel to the x-axis

Let us assume be the equation.

So, on substituting the values x1 = -2, y1 = 3, a = 3, and b = 2 in the equation, we have:

Thus, 4x2 + 9y2 + 16x – 54y + 61 = 0 is the required equation.

(ii) When the major axis is parallel to the y-axis

Let us assume be the equation.

So, on substituting the values x1 = -2, y1 = 3, a = 2, and b = 3 in the equation, we have:

Thus, 9x2 + 4y2 + 36x – 24y + 36 = 0 is the required equation.

### (i) Latus rectum is half of its minor axis

Solution:

Given that, 2b2/a = 2b/2

â‡’ 2b2 = ab

â‡’ 2b = a

Since,

â‡’

â‡’ e = âˆš3/2

Hence, the eccentricity of an ellipse is âˆš3/2

### (ii) Latus rectum is half of its major axis

Solution:

Given that, 2b2/a = 2a/2

â‡’ 2b2 = a2

Since,

â‡’

â‡’ e = 1/âˆš2

### (i) x2 + 2y2 – 2x + 12y + 10 = 0

Solution:

Given that x2 + 2y2 – 2x + 12y + 10 = 0

(x2 – 2x) + 2(y2 + 6y) = -10

â‡’ (x2 – 2x + 1) + 2(y2 + 6y + 9) = -10 + 18 + 1

â‡’

So, x1 = 1, y1 = -3

and a = 3  and b = 3/âˆš2

Centre = (1, -3)

Major axis = 2a = 2(3) = 6

Minor axis = 2b = 3\sqrt2

e=

= 1/âˆš2

Foci = (1 Â± 3/âˆš2; -3)

### (ii) x2 + 4y2 – 4x + 24y + 31 = 0

Solution:

Given that x2 + 4y2 – 4x + 24y + 31 = 0

(x2 – 4x) + 4(y2 + 6y) = -31

â‡’ (x2 – 4x + 4) + 4(y2 + 6y + 9) = 9

â‡’

So, x1 = 1, y1 = -3

and a = 3  and b = 3/2

Centre = (2, -3)

Major axis = 2a = 2(3) = 6

Minor axis = 2b = 3

e=

= âˆš3/2

Foci = (2 Â± 3/âˆš2; -3)

### (iii) 4x2 + y2 – 8x + 2y +1 = 0

Solution:

Given that 4x2 + y2 – 8x + 2y +1 = 0

4(x2 – 2x) + (y2 + 2y) = -1

4(x2 – 2x + 1) + (y2 + 2y + 1) = -1 + 4 + 1

4(x2 – 1) + (y2 + 1) = 4

â‡’

So, x1 = 1, y1 = -1

and a = 1  and b = 2

Centre = (1,-1)

Minor axis = 2a = 2(1) = 2

Minor axis = 2b = 4

e =

e = âˆš3/2

Foci = (1,  -1 Â± âˆš3)

### (iv) 3x2 + 4y2 – 12x – 8y + 4 = 0

Solution:

Given that 3x2 + 4y2 – 12x – 8y + 4 = 0

3(x2 – 4x) + 4(y2 – 2y) = -4

3(x2 – 4x + 4) + 4(y2 – 2y + 1) = -4 + 12 + 4

â‡’

So, x1 = 2, y1 = -1

and a = 2  and b = âˆš3

Centre = (2, 1)

Major axis = 2a = 2(2) = 4

Minor axis = 2b = 2(âˆš3) = 2âˆš3

e =

e = 1/2

Foci = (2 Â±  1, 1)

### (v) 4x2 + 16y2 – 24x – 32y – 12 = 0

Solution:

Given that 4x2 + 16y2 – 24x – 32y – 12 = 0

4(x2 – 6x) + 16(y2 – 2y) = 12

4(x2 – 6x + 9) + 16(y2 – 2y + 1) = 12 + 36 + 16

4(x – 3) + 16(y – 1) = 64

â‡’

So, x1 = 3, y1 = 1

and a = 4  and b = 2

Centre = (3, 1)

Major axis = 2a = 2(4) = 8

Minor axis = 2b = 2(2) = 4

e =

e = âˆš3/2

Foci = (3 Â±2âˆš3; 1)

### (vi) x2 + 4y2 – 2x = 0

Solution:

Given that x2 + 4y2 – 2x = 0

(x2 – 2x) + 4y2 = 0

(x2 – 2x + 1) + 4y2 = 0

(x – 1)2 + 4y2 = 0

â‡’

So, x1 = 1, y1 = 0

and a = 1  and b = 1/4

Centre = (1, 0)

Major axis = 2a = 2(1) = 2

Minor axis = 2b = 2(1/2) = 1

e =

e = âˆš3/2

Foci = (1 Â±âˆš3/2, 0)

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