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

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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