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

### 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[x

^{2 }+ 1 – 2x + y^{2 }+ 4 + 4y] =â‡’ 52[x

^{2 }+ 1 – 2x + y^{2 }+ 4 + 4y] = (3x – 2y + 5)^{2}â‡’ 52[x

^{2 }+ 1 – 2x + y^{2 }+ 4 + 4y] = 9x^{2 }+ 4y^{2}+ 25 – 12xy – 20y + 30x

Thus 43x^{2}+ 43y^{2}+ 12xy â€“ 134x + 228y + 235 = 0 is the required equation.

### Question 2. Find the equation of the ellipse if:

### (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[x

^{2}+ y^{2 }– 2y + 1] = x^{2 }+ y^{2 }+ 2xy

Thus 7x^{2}+ 7y^{2}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}/2â‡’ 8[(x + 1)

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

^{2 }+ (y – 1)^{2}] = x^{2}+ y^{2}+ 9 – 6y – 2xy + 6xâ‡’ 8[(x

^{2}+ 1 + 2x)^{ }+ (y^{2}+ 1 – 2y)] = x^{2}+ y^{2}+ 9 – 6y – 2xy + 6xâ‡’ 8[x

^{2}+ y^{2 }+ 2 + 2x – 2y] = x^{2}+ y^{2}+ 9 – 6y – 2xy + 6x

Thus 7x^{2}+ 7y^{2}+ 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[x^{2 }+ y^{2 }+ 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}/25â‡’ 100[(x – 1)

^{2}+ (y – 2)^{2}] = 9x^{2}+ 16y^{2}+ 25 + 24xy – 40y – 30xâ‡’ 100[(x

^{2}+ 1 – 2x) + (y^{2}+ 4 – 4y)] = 9x^{2}+ 16y^{2}+ 25 + 24xy – 40y – 30x

Thus 91x^{2}+ 84y^{2}â€“ 24xy â€“ 170x â€“ 360y + 475 = 0 is the required equation.

### Question 3. Find the eccentricity, coordinates of foci, length of the latus- rectum of the ellipse:

### (i) 4x^{2} + 9y^{2} = 1

**Solution:**

Given that 4x

^{2}+ 9y^{2}= 1So,

â‡’ Eccentricity =

= âˆš5/3

Length of latus rectum =

= 4/9

Foci are(âˆš5/6; 0) and (-âˆš5/6; 0)

### (ii) 5x^{2} + 4y^{2} = 1

**Solution:**

Given that 5x

^{2}+ 4y^{2}= 1So,

â‡’ Eccentricity =

= 1/âˆš5

Length of latus rectum =

= 4/5

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

### (iii) 4x^{2} + 3y^{2} = 1

**Solution:**

Given that 4x

^{2}+ 3y^{2}= 1So,

â‡’ Eccentricity =

= 1/2

Length of latus rectum = 2a

^{2}/b=

= âˆš3/2

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

### (iv) 25x^{2} + 16y^{2} = 1600

**Solution:**

Given that 25x

^{2}+ 16y^{2}= 1600â‡’

So,

â‡’ Eccentricity =

= 3/5

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

â‡’ Length of latus rectum = 2a

^{2}/b= 2 x (64/10)

= 64/5

### (v) 9x^{2} + 25y^{2} = 225

**Solution:**

Given that 9x

^{2}+ 25y^{2}= 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 = 2b

^{2}/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‡’ b

^{2}/a^{2}= 3/5â‡’ b

^{2 }= 3a^{2}/5â‡’ b

^{2 }= 3a^{2}/5 ……(iii)Now put the value of b

^{2}in equation (ii), we get9 + 5/3 = a

^{2}a

^{2 }= 32/3Now put the value of a

^{2 }in eq(iii), we get,b

^{2 }= 3/5 x 32/3 = 32/5Now put the a

^{2 }and b^{2 }in eq(i), we get,

Thus 3x^{2}+ 5y^{2}= 32 is the required equation of the plane.

### Question 5. Find the equation of the ellipse if:

### (i) e = 1/2 and foci (Â±2, 0).

**Solution:**

Let the equation of the ellipse be:

…(i)

Now, ae = 2

or, a

^{2 }= 16Now, b

^{2}= a^{2}(1 â€“ e^{2})â‡’ b

^{2}= 16(1 â€“ (1/2)^{2})â‡’ b

^{2}= 12Now put the a

^{2 }and b^{2 }in eq(i), we get,

Thus 3x^{2}+ 4y^{2}= 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‡’ 2b

^{2}/a = 5â‡’ b

^{2 }= 5a/2 ….(ii)Since, b

^{2 }= a^{2}(1 – e^{2})â‡’ 5a/2

^{ }= a^{2}(1 – (2/3)^{2})â‡’ a = 9/2

â‡’ a

^{2 }= 81/4Now put the value of a in eq(ii), we get

b

^{2 }= 5/2 x 9/2b

^{2 }= 45/4Now put the a

^{2 }and b^{2 }in eq(i), we get,

Thus 20x^{2}+ 36y^{2}= 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‡’ a

^{2}= 16We know, a

^{2 }= b^{2}(1 – e^{2})â‡’ 16 = b

^{2}(1 – (1^{2}/2^{2}))â‡’ b

^{2}= 12Now put the a

^{2 }and b^{2 }in eq(i), we get,â‡’

Thus 3x^{2}+ 4y^{2}= 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‡’ b

^{2 }= 27On substituting the values of a

^{2}and b^{2 }in eq(i), we get,â‡’

â‡’

Thus 3x^{2}+ 4y^{2}= 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/a

^{2 }and r = 1/b^{2}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 3x^{2}+ 7y^{2}= 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, b

^{2 }= a^{2}(1 – e^{2})â‡’ b

^{2}= 25(1 – 16/25)â‡’ b

^{2 }= 9On substituting the values of a

^{2}and b^{2 }in eq(i), we get,

Thusis 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, a

^{2 }= b^{2}(1 – e^{2})â‡’ a

^{2 }= 169(1 – 25/169)â‡’ a

^{2}= 144On substituting the values of a

^{2}and b^{2 }in eq(i), we get,

Thusis 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, b

^{2 }= a^{2}(1 – e^{2})â‡’ b

^{2 }= 36(1 – 16/36)â‡’ b

^{2 }= 20On substituting the values of a

^{2}and b^{2 }in eq(i), we get,

Thusis 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, a

^{2 }= 9, b^{2 }= 4On substituting the values of a

^{2}and b^{2 }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, a

^{2 }= 1, b^{2 }= 5On substituting the values of a

^{2}and b^{2 }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‡’ a

^{2}= 169Also, ae = 5

â‡’ e = 5/13

We know,

â‡’

â‡’ b

^{2}= 144On substituting the values of a

^{2}and b^{2 }in eq(i), we get,

Thusis 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

a

^{2}= 64Now the coordinates of foci are (0, Â±be)

So, be = 6

(be)

^{2}= 36We know, a

^{2 }= b^{2}(1 – e^{2})a

^{2 }= b^{2}– b^{2}e^{2}64

^{ }= b^{2}– 36b

^{2 }= 100On substituting the values of a

^{2}and b^{2 }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 a

^{2}= 16We know,

â‡’

â‡’ b

^{2}= 7On substituting the values of a

^{2}and b^{2 }in eq(i), we get,

Thusis 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 a

^{2 }= 144We know that, b

^{2 }= a^{2}(1 – e^{2})â‡’ b

^{2 }= 144(1 – (1/3)^{2})â‡’ b

^{2 }= 144(8/9)â‡’ b

^{2}= 128On substituting the values of a

^{2}and b^{2 }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‡’ b

^{2}= a^{2}e^{2 }….(i)Also given that the latus rectum is 10

So, 2b

^{2}/a = 10b

^{2}= 5a ….(ii)As we know that b

^{2 }= a^{2}(1 – e^{2})â‡’ b

^{2 }= a^{2}(1 – e^{2})â‡’ b

^{2 }= a^{2 }– a^{2}e^{2}â‡’ b

^{2 }= a^{2 }– b^{2}â‡’ 2b

^{2 }= a^{2}â‡’ b

^{2 }= a^{2}/2 ….(iii)Now put the value of b

^{2 }in eq(ii), we geta

^{2}/2 = 5aa = 10

So, a

^{2}= 100Now put the value a

^{2 }of in eq(iii), we getb

^{2 }= 100/2b

^{2 }= 50On substituting the values of a

^{2}and b^{2 }in eq(i), we get,

Hence, x^{2}+ 2y^{2}= 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-axisLet us assume be the equation.

So, on substituting the values x

_{1 }= -2, y_{1 }= 3, a = 3, and b = 2 in the equation, we have:

Thus, 4x^{2}+ 9y^{2}+ 16x – 54y + 61 = 0 is the required equation.

(ii)When the major axis is parallel to the y-axisLet us assume be the equation.

So, on substituting the values x

_{1 }= -2, y_{1 }= 3, a = 2, and b = 3 in the equation, we have:

Thus, 9x^{2}+ 4y^{2}+ 36x – 24y + 36 = 0 is the required equation.

### Question 9. Find the eccentricity of an ellipse whose:

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

**Solution:**

Given that, 2b

^{2}/a = 2b/2â‡’ 2b

^{2 }= abâ‡’ 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, 2b

^{2}/a = 2a/2â‡’ 2b

^{2 }= a^{2}Since,

â‡’

â‡’ e = 1/âˆš2

### Question 10. Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse:

### (i) x^{2} + 2y^{2} – 2x + 12y + 10 = 0

**Solution:**

Given that x

^{2}+ 2y^{2}– 2x + 12y + 10 = 0(x

^{2}– 2x) + 2(y^{2 }+ 6y) = -10â‡’ (x

^{2}– 2x + 1) + 2(y^{2}+ 6y + 9) = -10 + 18 + 1â‡’

So, x

_{1 }= 1, y_{1}= -3and 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) x^{2} + 4y^{2} – 4x + 24y + 31 = 0

**Solution:**

Given that x

^{2}+ 4y^{2}– 4x + 24y + 31 = 0(x

^{2}– 4x) + 4(y^{2}+ 6y) = -31â‡’ (x

^{2}– 4x + 4) + 4(y^{2}+ 6y + 9) = 9â‡’

So, x

_{1}= 1, y_{1}= -3and 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) 4x^{2} + y^{2} – 8x + 2y +1 = 0

**Solution:**

Given that 4x

^{2}+ y^{2}– 8x + 2y +1 = 04(x

^{2}– 2x) + (y^{2}+ 2y) = -14(x

^{2}– 2x + 1) + (y^{2}+ 2y + 1) = -1 + 4 + 14(x

^{2}– 1) + (y^{2}+ 1) = 4â‡’

So, x

_{1}= 1, y_{1}= -1and 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) 3x^{2} + 4y^{2} – 12x – 8y + 4 = 0

**Solution:**

Given that 3x

^{2}+ 4y^{2}– 12x – 8y + 4 = 03(x

^{2}– 4x) + 4(y^{2}– 2y) = -43(x

^{2}– 4x + 4) + 4(y^{2}– 2y + 1) = -4 + 12 + 4â‡’

So, x

_{1}= 2, y_{1}= -1and 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) 4x^{2} + 16y^{2} – 24x – 32y – 12 = 0

**Solution:**

Given that 4x

^{2}+ 16y^{2}– 24x – 32y – 12 = 04(x

^{2}– 6x) + 16(y^{2}– 2y) = 124(x

^{2}– 6x + 9) + 16(y^{2}– 2y + 1) = 12 + 36 + 164(x – 3) + 16(y – 1) = 64

â‡’

So, x

_{1}= 3, y_{1}= 1and 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) x^{2} + 4y^{2} – 2x = 0

**Solution:**

Given that x

^{2}+ 4y^{2}– 2x = 0(x

^{2}– 2x) + 4y^{2}= 0(x

^{2}– 2x + 1) + 4y^{2}= 0(x – 1)

^{2}+ 4y^{2}= 0â‡’

So, x

_{1}= 1, y_{1}= 0and 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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