Hyperbola Exercise 8.2 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to Study

1. The Standard equation to a Hyperbola
   

Exercise 8.2

1. Find the coordinates of the vertices, the eccentricity, and the coordinates of the foci of the hyperbola.

a. $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{4} = 1$

Solution

a² = 16, b² = 4

So, a = 4, b = 2.

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

Or, e² = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = $\frac{{16 + 4}}{{16}}$ = $\frac{5}{4}$.

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

The coordinates of the foci = (±ae,0) = $\left( { \pm 4{\rm{*}}\frac{{\sqrt 5 }}{2},0} \right)$ = (±2$\sqrt 5 $, 0).

 

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

Solution

Or, $\frac{{{{\rm{y}}^2}}}{{16}} - \frac{{{{\rm{x}}^2}}}{9}$ = 1

Transverse axis is along the y – axis.

a² = 16, b² = 9.

So, a = 4, b = 3.

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

Or, e² = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = $\frac{{16 + 9}}{{16}}$ = $\frac{{25}}{{16}}$.

So, e = $\frac{5}{4}$.

The coordinates of the foci = (0,±ae) = $\left( {0, \pm 4{\rm{*}}\frac{5}{4}} \right)$ = (0,±5).

 

c. 3x² – 4y² = 36

Solution

3x² – 4y² = 36.

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

Or, a² = 12, b²=  9.

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

The coordinates of the vertices = (±a,0) = $\left( { \pm 2\sqrt 3 ,0} \right)$.

e² = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{{{{\rm{a}}^2}}}$ = $\frac{{12 + 9}}{{12}}$ = $\frac{{21}}{{12}}$ = $\frac{7}{4}$.

So, e = $\frac{{\sqrt 7 }}{2}$.

The coordinates of the foci = (±ae,0) = $\left( { \pm 2\sqrt 3 {\rm{*}}\frac{{\sqrt 7 }}{2},0} \right)$ = (±$\sqrt {21} $,0).

 

d. x² – 4y² – 4x = 0

Solution

Given hyperbola is x² – 4y² – 4x = 0

Or, (x – 2)² – 4y² = 4

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

So, a² = 4, b² = 1.

Centre (h,k) = (2,0)

We have, b² = a²(e² – 1).

Or, $\frac{1}{4}$ = e² – 1.

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

Vertices = (h ± a,k) = (2 ± 2,0) = (4,0),(0,0)

And foci = (h ± ae,k) = $\left( {2 \pm 2.\frac{{\sqrt 5 }}{2},0} \right)$ = (2 ± $\sqrt 5 $,0).

 

e. 9x² + 36x – 16y² + 32y = 124

Solution

Given hyperbola is, 9x² + 36x – 16y² + 32y = 124.

Or, 9(x + 2)² – 16(y – 1)² = 36 – 16 + 124 = 144.

So, $\frac{{{{\left( {{\rm{x}} + 2} \right)}^2}}}{{16}} - \frac{{{{\left( {{\rm{y}} - 1} \right)}^2}}}{9}$ = 1 …(i)

So, centre(h,k) = (- 2,1),a² = 16, b² = 9.

Vertices = (h ± a,k) = (- 2 ± 4, 1) = (2,1),(-6,1).

Eccentricity, e = $\sqrt {1 + \frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} $ = $\sqrt {1 + \frac{9}{{16}}} $ = $\frac{5}{4}$.

And, foci = (h ± ae,k) = $\left( { - 2 \pm 4.{\rm{\: }}\frac{5}{4},1} \right)$ = (3,1), (-7,1).

 f. 16x²-9y²-64x-54y+127=0

Solution:

16x² - 64x - 9y² - 54y + 127 = 0

Or, 16(x² - 4x) - 9(y² + 6y) + 127 = 0

Or, 16(x² - 4x + 4) -16(4) - 9(y² + 6y + 9) + 9(9) + 127 = 0

Or, 16(x - 2)² - 9(y + 3)² = -144

Or, $ \Rightarrow \frac{{16{{(x - 2)}^2} - 9{{(y + 3)}^2}}}{{144}} =  - 1$

Or, $ \Rightarrow \frac{{{{(x - 2)}^2}}}{9} - \frac{{{{(y + 3)}^2}}}{{16}} =  - 1$

Comparing with $\frac{{{{(x - h)}^2}}}{{{a^2}}} - \frac{{{{(y - k)}^2}}}{{{b^2}}} = -1$

a=3, b=4, h=2, k=3

Thus,

Vertices = (h,k±b) = (2,-3±4) = (2,1) or (2,-7)

Eccentricity = $\sqrt {1 + \frac{{{a^2}}}{{{b^2}}}} $

$ = \sqrt {1 + \frac{{{3^2}}}{{{4^2}}}}  = \frac{5}{4}$

Foci = (h,k±be)= (2,-3±4×(5/4)) = (2,2) or (2,-8)

2. Deduce the equation to the hyperbola in the standard form.

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

Solution

Vertex at (2,0) and focus at (-5,0),

Here, a = 2, ae = 5

Or, ae = 5.

Or, 2.e = 5

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

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

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

Or, $\frac{{{{\rm{x}}^2}}}{4} - \frac{{{{\rm{y}}^2}}}{{21}}$ = 1

Or, 21x² – 4y² = 84.

 

b. with a focus at (0,5) and a vertex at (0,-3).

Solution
Transverse axis is along the y – axis. Focus at (0,5) and vertex (0.-3).

ae = 5 and a = 3.

Or, ae = 3.

Or, 3e = 5

So, e = $\frac{5}{3}$.

Or, b² = a²(e² – 1) = 9 $\left( {\frac{{25}}{9} - 1} \right)$ = 16.

The equation of the hyperbola is:

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

Or, $\frac{{{{\rm{y}}^2}}}{9}$ – $\frac{{{{\rm{x}}^2}}}{{16}}$ = 1.

So, 16y² – 9x² = 144.

 

c. with a focus at (-7,0) and eccentricity 7/4.

Solution

Or, ae = 7.

Or, e = $\frac{7}{4}$.

Or, a. $\frac{7}{4}$ = 7

So, a = 4.
or, b² = a²(e² – 1) = 16 $\left( {\frac{{49}}{{16}} - 1} \right)$ = 33

The equation of the hyperbola is

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

Or, $\frac{{{{\rm{x}}^2}}}{{16}} - \frac{{{{\rm{y}}^3}}}{{33}}$ = 1

 

d. with a vertex at (0,8) and passing through (4,8 $\sqrt 2 $)

Solution

Transverse axis along the y – axis.

Here, a = 8.

The equation of the hyperbola is:

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

Or, $\frac{{{{\rm{y}}^2}}}{{64}}$ – $\frac{{{{\rm{x}}^2}}}{{{{\rm{b}}^2}}}$ = 1 …(i)

It passes through (4,8$\sqrt 2 $),so.

Or, $\frac{{128}}{{64}} - \frac{{16}}{{{{\rm{b}}^2}}}$ = 1.

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

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

So, b² = 16.

From (i) $\frac{{{{\rm{y}}^2}}}{{64}}$ – $\frac{{{{\rm{x}}^2}}}{{16}}$ = 1.

Which is the required equation of the parabola.

 

e. with length of the transverse axis =8 and eccentricity =2

Given,

Length of transverse axis = 2a = 8 → a=4

Length of conjugate axis =2b

Eccentricity = 2

i.e., $\sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}  = 2$

Squaring Both sides,

$[1 + \frac{{{b^2}}}{{{a^2}}} = 4$

$\frac{{{a^2} + {b^2}}}{{{a^2}}} = 4$

Or, 3a²=b²

Or, 3×4²=b²

Or, b²=48

And a²=16

Thus, Equation of Hyperbola = $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$

$\Rightarrow \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{4} = 1$