11th NCERT/CBSE Introduction to Introduction to Conic section Exercise 11.4 Questions 15
Hi

Question (1)

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$

Solution

$\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$
Comparing to standard form of parabola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
a2 = 16 ⇒ a = 4
b2 = 9 ⇒ b = 3
c2 = a2 + b2
c2 =16 + 9 = 25
c = 5
coordinate of foci = (±c, 0) = (±5, 0)
coordinate of vertices =(±a, 0) = (±4, 0)
eccentricity $e = \frac{c}{a} = \frac{5}{4}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 9}}{4} = \frac{9}{2}$

Question (2)

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{{{y^2}}}{9} - \frac{{{x^2}}}{{27}} = 1$

Solution

$\frac{{{y^2}}}{9} - \frac{{{x^2}}}{{27}} = 1$
As it is form $\frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1$
It is along y-axis
Comparing to standard form b2 = 9, a2 = 27
c2 = a2 + b2
c2 = 9 + 27 = 36
c = 6
Co-ordinate of foci = (0, ±c) = (0, ±6)
Co-ordinate of vertices = (0, ±b) = (0, ±3)
eccentricity $e = \frac{c}{b} = \frac{6}{3} = 2$
Length of latus rectum $ = \frac{{2{a^2}}}{b} = 2 \times \frac{{27}}{3} = 18$

Question (3)

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36

Solution

the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36
$\frac{{9{y^2}}}{{36}} - \frac{{4{x^2}}}{{36}} = 1$
$\frac{{{y^2}}}{4} - \frac{{{x^2}}}{9} = 1$
Hyperbola is along y-axis
b2 = 4 ⇒ b = 2
a2 = p ⇒ a = 3
c2 = a2 + b2
c2 = 9+4 =13
c = √13
Coordinate of foci - (0, ±c) = (0, ±√13)
co-ordinate of vertices = (0, ±b) = (0, ±2)
eccentricity $e = \frac{c}{b} = \frac{{\sqrt {13} }}{2}$
Length of latus rectum $ = \frac{{2{a^2}}}{b} = \frac{{2 \times 9}}{2} = 9$

Question (4)

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x2 – 9y2 = 576

Solution

16x2 - 9y2 = 576
$\frac{{16{x^2}}}{{576}} - \frac{{9{y^2}}}{{576}} = 1$
$\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{64}} = 1$
a2 = 36 and b2 = 64
It is along x-axis
c2 = a2 + b2
c2 = 36 + 64 = 100
c = 10
coordinate of foci = (±c, 0) = (±10, 0)
coordinate of vertices = (±a, 0) = (±6, 0)
eccentricity $e = \frac{c}{a} = \frac{{10}}{6} = \frac{5}{3}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 64}}{6} = \frac{{64}}{3}$

Question (5)

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 – 9x2 = 36

Solution

5y2 – 9x2 = 36
$\frac{{5{y^2}}}{{36}} - \frac{{9{x^2}}}{{36}} = 1$
$\frac{{{y^2}}}{{\frac{{36}}{5}}} - \frac{{{x^2}}}{{\frac{{36}}{9}}} = 1$
$\frac{{{y^2}}}{{\frac{{36}}{5}}} - \frac{{{x^2}}}{4} = 1$
Hyperbole is along y-axis
${b^2} = \frac{{36}}{5},{a^2} = 4$
c2 = a2 + b2
${c^2} = 4 + \frac{{36}}{5} = \frac{{56}}{5}$
$ \Rightarrow c = \sqrt {\frac{{56}}{5}} $
coordinate of foci = (±c, 0) $ = \left( {0, \pm \frac{{\sqrt {56} }}{{\sqrt 5 }}} \right) = \left( {0, \pm \frac{{2\sqrt {14} }}{{\sqrt 5 }}} \right)$
eccentricity $e = \frac{c}{b} = \frac{{\frac{{\sqrt {56} }}{{\sqrt 5 }}}}{{\frac{6}{{\sqrt 5 }}}} = \frac{{\sqrt {56} }}{6}$
$e = \frac{{2\sqrt {14} }}{6} = \frac{{\sqrt {14} }}{3}$
Length of latus rectum $ = \frac{{2{a^2}}}{b} = \frac{{2 \times 4}}{{\frac{6}{{\sqrt 5 }}}} = \frac{{4\sqrt 5 }}{3}$

Question (6)

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y2 – 16x2 = 784

Solution

49y2 – 16x2 = 784
$\frac{{49{y^2}}}{{784}} - \frac{{16{x^2}}}{{784}} = 1$
$\frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{{49}} = 1$
It is along y-axis, b2 = 16, a2 = 49
c2 = a2 + b2 =16+49=65
c=√65
coordinate of foci = (±c, 0)=(0,±√65)
coordinates of vertices = (0, ±b) = (0, ±4)
eccentricity $e = \frac{c}{b} = \frac{{\sqrt {65} }}{4}$
LEngth of latus rectum $ = \frac{{2{a^2}}}{b} = \frac{{2 \times 49}}{4} = \frac{{49}}{2}$

Question (7)

Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)

Solution

Vertices (±2, 0), foci (±3, 0)
y-coordinate of foci is zero . So focus is on x-axis
so hyperbola is along x-axis
foci = (±c, 0) = (±3, 0) ⇒ c = 3
Vertices (±2, 0) = (±a, 0) ⇒ a = 2
c2 = a2 + b2
9 = 4 + b2
b2 = 5
Equation of hyperbola is $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$

Question (8)

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8)

Solution

Vertices (±2, 0), foci (±3, 0)
Since x-coordinate of foci is zero foci is on y-axis
So hyperbola is along y-axis
foci = (0, ±c) = (0, ±8) ⇒ c = 8
vertices = (0, ±b) = (0, ±5) ⇒ b = 5
c2 = a2 + b2
64 = 25 + a2
a2 = 64 -25 = 39
So equation of hyperbola is $\frac{{{y^2}}}{{25}} - \frac{{{x^2}}}{{39}} = 1$

Question (9)

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5)

Solution

Vertices (0, ±3), foci (0, ±5)
Since x-ordinate of foci is zero, foci are y-axis
. so hyperbole is along y-axis
foci = (0, ±c) = (0, ±5) ⇒ c = 5
vertices = (0, ±b) = (0, ±3) ⇒ b = 3
c2 = a2 + b2
25 = a2 + 9
a2 = 16
So equation of hyperbole is $\frac{{{y^2}}}{9} - \frac{{{x^2}}}{{16}} = 1$

Question (10)

Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.

Solution

Foci (±5, 0), the transverse axis is of length 8.
y-coordinate of foci is O. So foci are on x-axis
x-axis is called transverse axis
Length of transverse axes = 2a = 8 ⇒ a = 4
foci (±c, 0) = (±5, 0) ⇒ c = 5
c2 = a2 + b2
25 = 16 + b2
b2=9
So equation of hyper bola is $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$

Question (11)

Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.

Solution

Foci (0, ±13), the conjugate axis is of length 24.
x-cordinate of foci is zero , so foci are on y-axis
Hyperbola is along x-axis . so x aixis conjugate axes
foci = (0, ±c) = (a, ±13) ⇒ c = 13
Length of conjugate axes = 24
2a = 24 ⇒ a = 12
c2 = a2 + b2
169 = 144 + b2
b2 = 25
Equation of hyperbola is
$\frac{{{y^2}}}{{25}} - \frac{{{x^2}}}{{144}} = 1$

Question (12)

Find the equation of the hyperbola satisfying the give conditions: Foci $\left( { \pm 3\sqrt 5 ,0} \right)$, the latus rectum is of length 8.

Solution

Foci $\left( { \pm 3\sqrt 5 ,0} \right)$, the latus rectum is of length 8.
y-coordinate of foci iis zero. So focus liies on x-axis. So hyperbola is along x-axis
foci = (±c, 0) = (±3√5, 0) ⇒ c = 3√3
Length of latus rectum $ = \frac{{2{b^2}}}{a}$
$8 = \frac{{2{b^2}}}{a} \Rightarrow {b^2} = 4a$
c2 = a2 + b2
45 = a2 + 4a
a2 + 5a - 45 = 0
(a+9)(a-5) = 0
a = -9 or a = 5
a =-9 is not possible
a = 5 ⇒ a2 = 25
b2 = 4(5) = 20
Equation of hyperbola is
$\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{20}} = 1$

Question (13)

Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12

Solution

Foci (±4, 0), the latus rectum is of length 12
y-coordinate of foci is zero, foci are on y-axis, hyperbola is along y-axis
foci = (±c, 0) = (±4, 0) ⇒ c = 4
Length of latus rectum = 12
$\therefore \frac{{2{b^2}}}{a} = 12 \Rightarrow {b^2} = 6a$
c2 = a2 + b2
16 = a2 + 6a
a2 + 6a - 16 = 0
(a+8) (a-2) = 0
a = -8 or a = 2
But a > 0 ⇒ a≠ - 8 ∴ a = 2
∴ b2 = 6a = 12
Equation of hyperbola is
$\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1$

Question (14)

Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), $e = \frac{4}{3}$

Solution

Vertices (±7, 0), $e = \frac{4}{3}$
Since y-coordinate of vertices 0, so vertices are on x-axis . So hyperbola is along x-axis
vertices = (±a, 0) = (±7, 0) ⇒ a = 7
$e = \frac{c}{a}$
$\frac{4}{3} = \frac{c}{7} \Rightarrow c = \frac{{28}}{3}$
Now c2 = a2 + b2
${\left( {\frac{{28}}{3}} \right)^2} = 49 + {b^2}$
${b^2} = \frac{{784}}{9} - 49 = \frac{{784 - 441}}{9}$
$ \Rightarrow {b^2} = \frac{{343}}{9}$ Equation of hyperbola is
$\frac{{{x^2}}}{{49}} - \frac{{9{y^2}}}{{343}} = 1$

Question (15)

Find the equation of the hyperbola satisfying the give conditions: Foci $\left( {0, \pm \sqrt {10} } \right)$, passing through (2, 3)

Solution

Foci $\left( {0, \pm \sqrt {10} } \right)$, passing through (2, 3).
x-coordinate of foci is zero , foci are on y-axis
So hyperbola is along y-axis equation is
$\frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1$
It passes through (2, 3)
$\frac{9}{{{a^2}}} - \frac{4}{{{b^2}}} = 1$
$\frac{9}{{10 - {b^2}}} - \frac{4}{{{b^2}}} = 1$
9b2 - 4(10-b2) = b2(10-b2)
9b2 -40 + 4b2 = 10b2 - b4
b4 + 3b2 - 40 = 0
(b2 + 8) (b2 - 5) = 0
b2 = -8 (which is not possible) or b2 = 5
∴ b2 = 5
∴ a2 = 10 - b2 = 10 - 5 = 5
Equation of hyperbola is
$\frac{{{y^2}}}{5} - \frac{{{x^2}}}{5} = 1$
y2 - x2 = 5
Exercise11.3⇐
⇒Exercise11.5