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

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  • Last Updated : 01 Feb, 2023
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Question 1. Find the equation of the ellipse whose focus is (1,–2) and directrix is 3x – 2y + 5 = 0, and eccentricity is 1/2.

Solution:

Given that, 

Focus is (1, -2)

directrix is 3x – 2y + 5 = 0, 

eccentricity(e) is 1/2.

As we know, SP = ePM

⇒ SP = PM/2

⇒ (SP)2 = 1/4(PM)2 

⇒ 4(SP)2 = (PM)2

⇒ 4[(x – 1)2 + (y + 2)2] = [\frac{3x-2y+5}{\sqrt{3^2+2^2}}]^2

⇒ 4[x2 + 1 – 2x + y2 + 4 + 4y] = \frac{(3x-2y+5)^2}{13}

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

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

Thus 43x2 + 43y2 + 12xy – 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

⇒ SP = PM/2

⇒ (SP)2 = 1/4(PM)2

⇒ 4(SP)2 = (PM)2

⇒ 4[(x – 0)2 + (y – 1)2] = [\frac{x+y}{\sqrt{1^2+1^2}}]^2

⇒ 8[x2 + y2 – 2y + 1] = x2 + y2 + 2xy

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

(ii) Focus is (–1,1), directrix is x – 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

⇒ SP = PM/2

⇒ (SP)2 = 1/4(PM)2

⇒ 4(SP)2 = (PM)2

⇒ 4[(x + 1)2 + (y – 1)2] = [\frac{x-y+3}{\sqrt{1^2+1^2}}]^2

⇒ 4[(x + 1)2 + (y – 1)2] = (x – y + 3)2/2 

⇒ 8[(x + 1)2 + (y – 1)2] = (x – y + 3)2

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

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

⇒ 8[x2 + y2 +  2 + 2x – 2y] = x2 + y2 + 9 – 6y – 2xy + 6x

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

(iii) Focus is (–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

⇒ SP = 4/5(PM) 

⇒ (SP)2 = 16/25(PM)2 

⇒ 25(SP)2 = 16(PM)2

⇒ 25[(x + 2)2 + (y – 3)2] = 16[\frac{2x+3y+4}{\sqrt{2^2+3^2}}]^2

⇒ 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 – 5 = 0 and e = 1/2.

Solution:

Given that,

focus is (1, 2), 

directrix is 3x + 4y – 5 = 0

eccentricity(e) is 1/2

We know, SP = ePM

⇒ SP = PM/2

⇒ (SP)2 = 1/4(PM)2

⇒ 4(SP)2 = (PM)2

⇒ 4[(x – 1)2 + (y – 2)2] = [\frac{3x+4y-5}{\sqrt{3^2+4^2}}]^2

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

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

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

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

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

(i) 4x2 + 9y2 = 1

Solution:

Given that 4x2 + 9y2 = 1

\frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{9}}=1

So, 

⇒ Eccentricity = \sqrt{\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{4}}}

= √5/3         

Length of latus rectum = \frac{\frac{2}{9}}{\frac{1}{2}}

= 4/9

Foci are(√5/6; 0) and (-√5/6; 0)  

(ii) 5x2 + 4y2 = 1

Solution:

Given that 5x2 + 4y2 = 1

\frac{x^2}{\frac{1}{5}}+\frac{y^2}{\frac{1}{4}}=1

So, 

⇒ Eccentricity = \sqrt{\frac{\frac{1}{4}-\frac{1}{5}}{\frac{1}{4}}}

= 1/√5 

Length of latus rectum = \frac{\frac{2}{5}}{\frac{1}{2}}

= 4/5

Foci are (0; 1/2√5) and (0; -1/2√5) 

(iii) 4x2 + 3y2 = 1

Solution:

Given that 4x2 + 3y2 = 1

\frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{3}}=1

So, 

⇒ Eccentricity = \sqrt{1-\frac{\frac{1}{4}}{\frac{1}{3}}}

= 1/2

Length of latus rectum = 2a2/b

\frac{\frac{2}{4}}{\frac{1}{\sqrt3}}

= √3/2 

Foci are (0; 1/2√3)  and (0; -1/2√3).        

(iv) 25x2 + 16y2 = 1600

Solution:

Given that 25x2 + 16y2 = 1600

\frac{25x^2}{1600}+\frac{16y^2}{1600}=1

⇒ \frac{25x^2}{1600}+\frac{16y^2}{1600}=1

So, 

⇒ Eccentricity = \sqrt{1-\frac{a^2}{b^2}}

= 3/5

⇒ Coordinates of foci are (0, 6) and (0, –6).

⇒ 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

=> \frac{x^2}{25}+\frac{y^2}{9}=1

Clearly, a = 5 and b = 3.

So,

⇒ Eccentricity = \sqrt{1-\frac{3^2}{5^2}}

= 4/5

⇒ Coordinates of foci are (4, 0) and (–4, 0).

⇒ Length of latus rectum = 2b2/a

= 2 x (9/5)

= 18/5

Question 4. Find the equation of the ellipse passing through the point (–3, 1) and has eccentricity \sqrt{\frac{2}{5}}     .      

Solution:

Let the equation of the plane be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       …(i)

It is given that the ellipse pass through the point (–3, 1), so,

\frac{9}{a^2}+\frac{1}{b^2}=1      …(ii)

⇒ e=\sqrt{1-\frac{a^2}{b^2}}

⇒ \sqrt{\frac{2}{5}}=\sqrt{1-\frac{a^2}{b^2}}

⇒ b2/a2 = 3/5 

⇒ b2 = 3a2/5

⇒ b2 = 3a2/5 ……(iii)

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

\frac{9}{a^2}+\frac{1}{\frac{3a^2}{5}}=1

\frac{9}{a^2}+\frac{5}{3a^2}=1

\frac{1}{a^2}[9 + \frac{5}{3}]=1

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,

\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1

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

Question 5. Find the equation of the ellipse if:

(i) e = 1/2 and foci (±2, 0).

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       …(i)

Now, ae = 2

or, a2 = 16

Now, b2 = a2(1 – e2)

⇒ b2 = 16(1 – (1/2)2)

⇒ b2 = 12

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

\frac{x^2}{16}+\frac{y^2}{12}=1

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:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       …(i)

Now, 

⇒ 2b2/a = 5

⇒ b2 = 5a/2  ….(ii)

Since, b2 = a2(1 – e2)

⇒ 5a/2  = a2(1 – (2/3)2)

⇒ a = 9/2

⇒ 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,

\frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1

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:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

Semi – major axis = a = 4

⇒ a2 = 16

We know, a2 = b2(1 – e2)

⇒ 16 = b2(1 – (12/22)) 

⇒ b2 = 12

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

⇒ \frac{x^2}{16}+\frac{y^2}{12}=1

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

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

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

Given, 2a = 12

⇒ a = 6

We know, b=\sqrt{1-\frac{b^2}{a^2}}

⇒ \frac{1}{2}=\sqrt{1-\frac{b^2}{36}}

⇒ b2 = 27

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

⇒ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1

⇒ \frac{x^2}{36}+\frac{y^2}{27}=1

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

(v) It passes through (1, 4) and (–6, 1).

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

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

\frac{1}{a^2}+\frac{16}{b^2}=1

 Let p = 1/a2 and r = 1/b2 

⇒ p + 16r = 1       ……(ii)

 Since the ellipse also passes through the point (–6, 1), we have

\frac{36}{a^2}+\frac{1}{b^2}=1

⇒ 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;

⇒ \frac{3x^2}{115}+\frac{7y^2}{115}=1

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

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

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

Given that a = 5 and ae = 4

Thus, e = 4/5 e=\frac{4}{5}

Now, b2 = a2(1 – e2)

⇒ b2 = 25(1 – 16/25) 

⇒ b2 = 9

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

Thus \frac{x^2}{25}+\frac{y^2}{9}=1       is the required equation.

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

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

Given: b = 13 and be = 5

Hence, e = 5/13 

Now, a2 = b2(1 – e2)

⇒ a2 = 169(1 – 25/169)

⇒ a2 = 144

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

\frac{x^2}{144}+\frac{y^2}{169}=1

Thus \frac{x^2}{144}+\frac{y^2}{169}=1       is the required equation.

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

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

Given: a = 6 and ae = 4

Thus, e = 2/3 

Now, b2 = a2(1 – e2)

⇒ b2 = 36(1 – 16/36)

⇒ b2 = 20

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

\frac{x^2}{36}+\frac{y^2}{20}=1

Thus \frac{x^2}{36}+\frac{y^2}{20}=1      is the required equation.

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

Solution:

Let the equation of the ellipse be \frac{x^2}{a^2}+\frac{y^2}{b^2}=1             …..(i)

Ends of major axis = (±3, 0)

Ends of minor axis = (0, ±2)

Since the ends of the major and minor axes are (±a, 0) and (0, ±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,

\frac{x^2}{9}+\frac{y^2}{4}=1

Thus, \frac{x^2}{9}+\frac{y^2}{4}=1             is the required equation.

(x) Ends of the major axis (0, ±√5) and ends of the minor axis (±1, 0).

Solution:

Let the equation of the ellipse be \frac{x^2}{a^2}+\frac{y^2}{b^2}=1              ….(i)

Ends of major axis = (0, ±√5)

Ends of minor axis = (±1, 0)

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

Hence, a = 1 and b = √5  

So, a2 = 1, b2 = 5

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

\frac{x^2}{1}+\frac{y^2}{5}=1

Thus, \frac{x^2}{1}+\frac{y^2}{5}=1      is the required equation.

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

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

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

We have, 2a = 26

⇒ a = 13

⇒ a2 = 169

 Also, ae = 5

⇒ e = 5/13 

We know, e=\sqrt{1-\frac{b^2}{a^2}}

⇒ \frac{5}{13}=\sqrt{1-\frac{b^2}{169}}

⇒ b2 = 144

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

\frac{x^2}{169}+\frac{y^2}{144}=1

Thus \frac{x^2}{169}+\frac{y^2}{144}=1       is the required equation.

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

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

Given that, length of minor axis is 16

2a = 16

a = 8

a2 = 64

Now the coordinates of foci are (0, ±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,

\frac{x^2}{64}+\frac{y^2}{100}=1

Thus, \frac{x^2}{64}+\frac{y^2}{100}=1       is the required equation.

(xiii) Foci are (±3, 0) and a = 4

Solution:

Let the equation of the ellipse be:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

Given that, ae = 3 and a = 4

Thus, e = 3/4 and a2 = 16

We know, e=\sqrt{1-\frac{b^2}{a^2}}

⇒ \frac{3}{4}=\sqrt{1-\frac{b^2}{16}}

⇒ b2 = 7

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

\frac{x^2}{16}+\frac{y^2}{7}=1

Thus \frac{x^2}{16}+\frac{y^2}{7}=1       is the required equation of the ellipse.

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

Solution:

Let the equation of the ellipse be \frac{x^2}{a^2}+\frac{y^2}{b^2}=1       ….(i)

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

We have, ae = 4 and  e = 1/3

Thus, a = 12

and a2 = 144

We know that, b2 = a2(1 – e2)

⇒ b2 = 144(1 – (1/3)2)

⇒ b2 = 144(8/9)

⇒ b2 = 128

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

\frac{x^2}{144}+\frac{y^2}{128}=1

Thus, \frac{x^2}{144}+\frac{y^2}{128}=1       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 (±ae, 0).

2b = 2ae 

⇒ b = ae

⇒ 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

⇒ b2 = a2(1 – e2

⇒ b2 = a2 – a2e2

⇒ b2 = a2 – b2

⇒ 2b2 = a2

⇒ 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,

\frac{x^2}{100}+\frac{y^2}{50}=1

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 \frac{(x-x_1)^2}{a^2}+\frac{(y-y_1)^2}{b^2}=1           be the equation.

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

\frac{(x+2)^2}{9}+\frac{(y-3)^2}{4}=1

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 \frac{(x-x_1)^2}{a^2}+\frac{(y-y_1)^2}{b^2}=1           be the equation.

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

\frac{(x+2)^2}{4}+\frac{(y-3)^2}{9}=1

Thus, 9x2 + 4y2 + 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, 2b2/a = 2b/2

⇒ 2b2 = ab

⇒ 2b = a

Since, e=\sqrt{1-\frac{b^2}{4a^2}}

⇒ e=\sqrt{1-\frac{1}{4}}

⇒ e = √3/2 

Hence, the eccentricity of an ellipse is √3/2 

(ii) Latus rectum is half of its major axis

Solution:

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

⇒ 2b2 = a2

Since, e=\sqrt{1-\frac{b^2}{2b^2}}

⇒ e=\sqrt{1-\frac{1}{2}}

⇒ e = 1/√2

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

(i) x2 + 2y2 – 2x + 12y + 10 = 0 

Solution:

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

(x2 – 2x) + 2(y2 + 6y) = -10

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

⇒ \frac{(x-1)^2}{9}+\frac{(y+3)^2}{\frac{9}{2}}=9

So, x1 = 1, y1 = -3

and a = 3  and b = 3/√2

Centre = (1, -3)

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

Minor axis = 2b = 3\sqrt2

e=\sqrt{1-\frac{b^2}{a^2}}

= 1/√2 

Foci = (1 ± 3/√2; -3) 

(ii) x2 + 4y2 – 4x + 24y + 31 = 0

Solution:

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

(x2 – 4x) + 4(y2 + 6y) = -31

⇒ (x2 – 4x + 4) + 4(y2 + 6y + 9) = 9

⇒ \frac{(x-1)^2}{9}+\frac{(y+3)^2}{\frac{9}{2}}=9

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= \sqrt{1-\frac{b^2}{a^2}}

= √3/2  

Foci = (2 ± 3/√2; -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

⇒ \frac{(x-1)^2}{1}+\frac{(y+1)^2}{4}=1

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 = \sqrt{1-\frac{b^2}{a^2}}

e = √3/2 

Foci = (1,  -1 ± √3) 

(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

⇒ \frac{(x-2)^2}{4}+\frac{(y-1)^2}{3}=1

So, x1 = 2, y1 = -1

and a = 2  and b = √3

Centre = (2, 1)

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

Minor axis = 2b = 2(√3) = 2√3

e = \sqrt{1-\frac{b^2}{a^2}}

e = 1/2  

Foci = (2 ±  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

⇒ \frac{(x-3)^2}{16}+\frac{(y-1)^2}{4}=1

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 = \sqrt{1-\frac{b^2}{a^2}}

e = √3/2  

Foci = (3 ±2√3; 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

⇒ \frac{(x-1)^2}{1}+\frac{(y)^2}{\frac{1}{4}}=9

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 = \sqrt{1-\frac{b^2}{a^2}}

e = √3/2 

Foci = (1 ±√3/2, 0) 


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