Ellipse Exercise 8.1 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to Study

1. The Standard equation to an Ellipse
2. Some of the terms of ellipse: Major Axis, Minor Axis, Eccentricity, Latus Rectum, Axis, Directrix
3. Equation of Ellipse with Centre not at origin
   

Exercise 8.1

1. Find the coordinates of the vertices, the eccentricity, the coordinates of the foci, the length of major and minor axes of the following ellipse.

(i). $\frac{{{{\rm{x}}^2}}}{{16}}$ + $\frac{{{{\rm{y}}^2}}}{4}$ = 1

Solution

Or, $\frac{{{{\rm{x}}^2}}}{{16}}$ + $\frac{{{{\rm{y}}^2}}}{4}$ = 1.

Also, $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}$ + $\frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1

Comparing, a² = 16, b² = 4.

SO, a = 4 , b = 2.

The coordinates of the vertices = (±a,0) = (±4,0)

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = 1 – $\frac{4}{{16}}$ = $\frac{3}{4}$.

So, e = $\frac{{\sqrt 3 }}{2}$.

The coordinates of the foci = (±ae,0) = (±4,$\frac{{\sqrt 3 }}{2},0)$ = (±$2\sqrt 3 $,0).

Length of major axis = 2a = 2 * 4 = 8.

Length of minor axis = 2b = 2 * 2 = 4.

 

(ii) $\frac{{{{\rm{x}}^2}}}{{10}}$ + $\frac{{{{\rm{y}}^2}}}{5}$ = 1

Solution

$\frac{{{{\rm{x}}^2}}}{{10}}$ + $\frac{{{{\rm{y}}^2}}}{5}$ = 1.

Also, $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}$ + $\frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1

a² = 10, b² = 5.

Comparing, a = $\sqrt {10} $ , b = $\sqrt 5 $.

The coordinates of the vertices = (±a,0) = (±$\sqrt {10} $,0)

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = 1 – $\frac{5}{{10}}$ = $\frac{1}{2}$.

So, e = $\frac{1}{{\sqrt 2 }}$.

The coordinates of the foci = (±ae,0) = (±$\sqrt {10} {\rm{\: }}$* $\frac{1}{{\sqrt 2 }}$, 0$)$ = (±$\sqrt 5 $,0).

Length of major axis = 2a = 2 * $\sqrt {10} $ = 2$\sqrt {10} $.

Length of minor axis = 2b = 2 $\sqrt 5 $.

 

(iii) $\frac{{{{\rm{x}}^2}}}{{9}}$ + $\frac{{{{\rm{y}}^2}}}{16}$ = 1

Solution

Here, the major axis is along y – axis.

$\frac{{{{\rm{x}}^2}}}{{9}}$ + $\frac{{{{\rm{y}}^2}}}{16}$ = 1.

Also, $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}$ + $\frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1

a² = 16, b² = 9.

Comparing, a = 4 , b = 3.

The coordinates of the vertices = (0,±a) = (0,±4)

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = 1 – $\frac{9}{{16}}$ = $\frac{7}{{16}}$.

So, e = $\frac{{\sqrt 7 }}{4}{\rm{\: }}$.

The coordinates of the foci = (0,±ae) = (0, ± 4 * $\frac{{\sqrt 7 }}{4}){\rm{\: }}$ =(0,±$\sqrt 7 $).

Length of major axis = 2a = 2 * 4 = 8.

Length of minor axis = 2b = 2 * 3 = 6.

 

(iv) 25x² + 4y² = 100

Solution

25x² + 4y² = 100

Or, $\frac{{{{\rm{x}}^2}}}{4}$ + $\frac{{{{\rm{y}}^2}}}{{25}}$ = 1.

Here, the major axis is along y – axis.

a² = 25, b² = 4.

So, a= 5 and b = 2.

The coordinates of the vertices = (0,±a) = (0,±5).

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = 1 – $\frac{4}{{25}}$ = $\frac{{21}}{{25}}$.

So, e = $\frac{{\sqrt {21} }}{5}$.

The coordinates of the foci = (o,±ae) = (0,±5 * $\frac{{\sqrt {21} }}{5}$) = (0,±$\sqrt {21} $)

Length of major axis = 2a = 2 * 5 = 10.

Length of minor axis = 2b = 2 * 2 = 4.

 

(v) 3x² + 4y² = 36

Solution

3x² + 4y² = 36

Or, $\frac{{{{\rm{x}}^2}}}{{12}}$ + $\frac{{{{\rm{y}}^2}}}{9}$ = 1.

Here, the major axis is along y – axis.

a² = 12, b² = 9.

So, a= 2$\sqrt 3 $ and b = 3.

The coordinates of the vertices = (±a,0) = (±2$\sqrt 3 $,0).

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = 1 – $\frac{9}{{12}}$ = $\frac{1}{4}$.

So, e = $\frac{1}{2}$.

The coordinates of the foci = (±ae,0) = (±2$\sqrt 3 $ * $\frac{1}{2}$, 0) = (±$\sqrt 3 $,0).

Length of major axis = 2a = 2 * 2$\sqrt 3 $ = 4$\sqrt 3 $.

Length of minor axis = 2b = 2 *3 = 6.

 

2. Deduce the equation of the ellipse in the standard position with the following data.

a. A focus at (-2,0) and a vertex at (5,0).

Solution

a = 5.

ac = 2.

5e = 2.

So, e = $\frac{2}{5}$.

b² = a²(1 – e²) = 25 $\left( {1 - \frac{4}{{25}}} \right)$ = 21.

The equation of the ellipse is:

Or, $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1.

Or, $\frac{{{{\rm{x}}^2}}}{{25}}$ + $\frac{{{{\rm{y}}^2}}}{{21}}$ = 1.

 

b. A vertex at (±9,0) and eccentricity 2/3.

Solution

ae = $\frac{{10}}{3}$, e = $\frac{2}{3}$.

So, $\frac{2}{3}$a = $\frac{{10}}{3}$

So, a = 5.

Or, b² = a²(1 – e²) = 25 $\left( {1 - \frac{4}{9}} \right)$ = $\frac{{125}}{9}$.

The equation of the ellipse is $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1.

Or, $\frac{{{{\rm{x}}^2}}}{{25}}$ + $\frac{{{{\rm{b}}^2}}}{{\frac{{125}}{9}}}$ = 1.

Or, $\frac{{{{\rm{x}}^2}}}{9}$ + $\frac{{9{{\rm{y}}^2}}}{{125}}$ = 1

Or, 5x² + 9y² = 125.

 

c. A focus at (0,-5) and eccentricity 1/3.

Solution
Major axis is along the y – axis.

ae = 5, e = $\frac{1}{3}$.

Or, a.$\frac{1}{3}$ = 5.

So, a = 15.

Or, b² = a²(1 – e²) = 225$\left( {1 - \frac{1}{9}} \right)$ = 200.

The equation of the ellipse is:

Or, $\frac{{{{\rm{x}}^2}}}{{200}}$ + $\frac{{{{\rm{y}}^2}}}{{225}}$ = 1.

Or, 225x² + 200y² = 45000

Or, 9x² + 8y² = 1800.

 

d. A focus at (0,3) and length of minor axis 8.

Solution

Major axis along the y –axis.

ae = 3, 2b = 8.

Or, b² = a²(1 – e²).

Or, b² = a² – a².e²

Or, 16 = a² – 9.

Or, a² = 16 + 9 = 25.

The equation of the ellipse is:

Or, $\frac{{{{\rm{x}}^2}}}{{16}}$ + $\frac{{{{\rm{y}}^2}}}{{25}}$ = 1.

 

e. A vertex at (0,8) and passing through (3,32/5)

Solution

Major axis is along the y – axis.

Here, a = 8.

The equation of the ellipse is $\frac{{{{\rm{x}}^2}}}{{{{\rm{b}}^2}}} + \frac{{{{\rm{y}}^2}}}{{64}}$ = 1…..(i)

Since it passes through $\left( {3,\frac{{32}}{5}} \right){\rm{\: }}$, so.

Or, $\frac{9}{{{{\rm{b}}^2}}}$ + $\frac{{1024}}{{25{\rm{*}}64}}$ = 1.

Or, $\frac{9}{{{{\rm{b}}^2}}} + \frac{{16}}{{25}}$ = 1.

Or, $\frac{9}{{{{\rm{b}}^2}}}$ = 1 –$\frac{{16}}{{25}}$

Or, $\frac{9}{{{{\rm{b}}^2}}}$ = $\frac{9}{{25}}$

So, b² = 25.

The equation of the ellipse is $\frac{{{{\rm{x}}^2}}}{{25}}$ + $\frac{{{{\rm{y}}^2}}}{{64}}$ = 1   [from(i)]

 

f. Passing through the points (1,4) and (-3,2).

Solution

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

It passes through (1,4) and (-3,2),so.

Or, $\frac{1}{{{{\rm{a}}^2}}}$ + $\frac{{16}}{{{{\rm{b}}^2}}}$ = 1

Or, $\frac{9}{{{{\rm{a}}^2}}}$ + $\frac{4}{{{{\rm{b}}^2}}}$ = 1.

Solving there two equations,

Or, a² = $\frac{{140}}{{12}}$ = $\frac{{35}}{3}$.

Or, b² = $\frac{{140}}{8}$ = $\frac{{35}}{2}$.

From (i), $\frac{{{{\rm{x}}^2}}}{{\frac{{35}}{3}}} + \frac{{{{\rm{y}}^2}}}{{\frac{{35}}{2}}}$ = 1.

Or, 3x² + 2y² = 35.

 

3. Find eccentricity, the coordinates of the vertices, Centre and the foci of the following ellipse.

a. $\frac{{{{(x + 2)}^2}}}{{16}} + \frac{{{{(y - 5)}^2}}}{9} = 1$

Solution
or, $\frac{{{{\left( {{\rm{x}} + 2} \right)}^2}}}{{16}}$ + $\frac{{{{\left( {{\rm{y}} - 5} \right)}^2}}}{9}$ = 1.

Also, $\frac{{{{\left( {{\rm{x}} - {\rm{h}}} \right)}^2}}}{{{{\rm{a}}^2}}}$ + $\frac{{{{\left( {{\rm{y}} - {\rm{k}}} \right)}^2}}}{{{{\rm{b}}^2}}}$ = 1.

Comparing h = - 2,k = 5, a² = 16, b² = 9.

So, a = 4, b = 3.

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = 1 – $\frac{9}{{16}}$ = $\frac{7}{{16}}$.

So, e = $\frac{{\sqrt 7 }}{4}$.

The coordinates of the vertices = (h ± a,k)

= (- 2 ± 4,5) i.e. (-6,5) and (2,5).

So, the coordinates of the centre = $\left( {\frac{{ - 6 + 2}}{2},\frac{{5 + 5}}{2}} \right)$ = (-2,5).

The coordinate of the foci = (h ± ae,k) = (- 2 ± 4 * $\frac{{\sqrt 7 }}{4}$,5)

= (- 2 ± $\sqrt 7 $, 5).

 

b. $\frac{{{{(x + 6)}^2}}}{4} + \frac{{{y^2}}}{{36}} = 1$

Solution

The major axis is along the y – axis.

y = - 6, k = 0, a² = 36, b² = 4.

So, a = 6, b = 2.

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = $1 - \frac{4}{{36}}$ = $\frac{8}{9}$.

So, e = $\frac{{2\sqrt 2 }}{3}$.

The coordinates of the vertices = (h,k ± a) = (-6,0 ± 6) = (-6,±6).

The coordinates of the centre = $\left( {\frac{{ - 6 - 6}}{2},\frac{{6 - 6}}{2}} \right)$.

The coordinates of the foci = (h,k ± ae).

= $\left( { - 6,0 \pm 6.\frac{{2\sqrt 2 }}{3}} \right)$ = (-6,± 4$\sqrt 2 $).

 

c.$ \frac{{{x^2}}}{8} + \frac{{{{(y - 2)}^2}}}{{12}} = 1$

Solution

Major axis is along the y – axis.

Here, a² = 12, b² = 8, h = 0, k = 2.

e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = 1 – $\frac{8}{{12}}$ = $\frac{1}{3}$.

So, e = $\frac{1}{{\sqrt 3 }}$.

The coordinates of the vertices = (h,k ± a) = (0,2 ± 2$\sqrt 3 $) = (0,2).

The coordinates of the foci = (h,k ± ae) = $\left( {0,2 \pm \sqrt 3 .\frac{1}{{\sqrt 3 }}} \right)$.

= (0,2 ± 2) ie (0,0) and (0,4).

 

d. 9x² + 5(y² – 6y) = 0

Solution

Given ellipse, is 9x² + 5(y² – 6y) = 0

Or, 9x² + 5(y – 3)² = 45.

So, $\frac{{{{\rm{x}}^2}}}{5} + \frac{{{{\left( {{\rm{y}} - 3} \right)}^2}}}{9}$ = 1…. (i)

So, a² = 5, b² = 9 (a < b)

Now, a² = b²(1 – e²).

Or, 5 = 9(1 – e²).

So, e = $\frac{2}{3}$.

Centre (h,k) = (0,3)

Foci = (h,k ± be) = $\left( {0,3 \pm \frac{{3.2}}{3}} \right)$ = (0,5),(0,1).

 

e. x² + 4y² – 4x + 24y + 24 = 0

Solution

The given ellipse is, x² + 4y² – 4x + 24y + 24 = 0.

Or, (x – 2)² + 4(y + 3)² = 4 + 36 – 24 = 16

So, $\frac{{{{\left( {{\rm{x}} - 2} \right)}^2}}}{{16}}$ + $\frac{{{{\left( {{\rm{y}} + 3} \right)}^2}}}{4}$ = 1 …(i)

So, a² = 16, b² = 4 and centre (h,k) = (2,-3).

We have, b² = a²(1 – e²).

So, e = $\frac{{\sqrt 3 }}{2}$.

And foci = (h ± ae,k) = $\left( {2 \pm 4.\frac{{\sqrt 3 }}{2}, - 3} \right)$ = (2 ± 2$\sqrt 3 $, - 3).

 

4. Find the equation of the ellipse:

a. major axis is twice its minor axis and which passes through the point (0,1)

Solution

The equation of the ellipse is:

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

By given, a = 2b

It passes through (0,1) so,

Or, 0 + $\frac{1}{{{{\rm{b}}^2}}}$ = 1

So, b² = 1

Or, a = 2b = 2 * 1 = 2.

From (i) $\frac{{{{\rm{x}}^2}}}{4} + \frac{{{{\rm{y}}^2}}}{1}$ = 1

So, x² + 4y² = 4.

 

b. latus rectum is equal to half its minor axis and which passes through th epoint ($\sqrt 6 $,1)

Solution

The equation of the ellipse is:

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

Latus rectum = $\frac{{2{{\rm{b}}^2}}}{{\rm{a}}}$.

Major axis = 2a.

By given, $\frac{{2{{\rm{b}}^2}}}{{\rm{a}}}$ = $\frac{1}{2}$.2a.

So, a² = 2b².

From (i) $\frac{{{{\rm{x}}^2}}}{{2{{\rm{b}}^2}}}$ + $\frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1.

It passes thorugh ($\sqrt 6 $,1), so $\frac{6}{{2{{\rm{b}}^2}}}$ + $\frac{1}{{{{\rm{b}}^2}}}$ = 1.

Or, $\frac{4}{{{{\rm{b}}^2}}}$ = 1.

So, b² = 4

Or, a² = 2b² = 2 * 4 = 8.

From (i) $\frac{{{{\rm{x}}^2}}}{8}$ + $\frac{{{{\rm{y}}^2}}}{4}$ = 1.

Or, x² + 2y² = 8.

 

c. latus rectum us 3 and ecentricity is $\frac{1}{{\sqrt 2 }}$

Solution

Or, $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}$ + $\frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1.

By given, $\frac{{2{{\rm{b}}^2}}}{{\rm{a}}}$ = 3.

So, b² = $\frac{{3{\rm{a}}}}{2}$.

Or, e² = 1 – $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$.

Or, $\frac{1}{2}$ = 1 – $\frac{{\frac{{3{\rm{a}}}}{2}}}{{{{\rm{a}}^2}}}$.

Or, $\frac{3}{{2{\rm{a}}}}$ = 1 – $\frac{1}{2}$ = $\frac{1}{2}$.

So, a = 3.

Or, b² = $\frac{3}{2}$ * 3 = $\frac{9}{2}$.

From (i) $\frac{{{{\rm{x}}^2}}}{9}$ + $\frac{{{{\rm{y}}^2}}}{{\frac{9}{2}}}{\rm{\: }}$ = 1.

So, x² + 2y² = 9.

 

d. Distance between the two foci is 8 and the semi latus rectum is 6.

Solution

Let the ellipse be $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}$ + $\frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1 …(1) (a > b).

So, distance between foci, 2ae = 8.

So, ae = 4 and semi – latus rectum $\frac{{{{\rm{b}}^2}}}{{\rm{a}}}$ = 6.

We have, b² = a²(1 – e²).

Or, 6a = a² – 16

Or, a² – 6a – 16 = 0

Or, a² – 8a + 2a – 16 = 0

So, a = 8, - 2. (a ≠ -2).

So, b² = 6a = 6 * 8 = 48.

Thus, from (i), $\frac{{{{\rm{x}}^2}}}{{64}}$ + $\frac{{{{\rm{y}}^2}}}{{48}}$ = 1.

So, 3x² + 4y² = 192.

 

e. latus rectum is half the major axis and focus is at (3,0)

Solution

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

So, focus (±ae,0) = (3,0)

So, ae = 3.

And $\frac{{2{{\rm{b}}^2}}}{{\rm{a}}}$ = $\frac{1}{2}$(2a).

Or, 2.a²(1 – e²) = a² à e = $\frac{1}{{\sqrt 2 }}$.

So, a = 3$\sqrt 2 $, b² = a²(1 – e²) = 9

Thus from (i), $\frac{{{{\rm{x}}^2}}}{{18}}$ + $\frac{{{{\rm{y}}^2}}}{9}$ = 1.

So, x² + 2y² = 18.