11th NCERT/CBSE Introduction to Introduction to Conic section Exercise 11.3 Questions 20

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Question (1)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{16}} = 1$

Solution

$\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{16}} = 1$
Comparing to standard form
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ we get
a² = 36, b² = 16
⇒ a = 6, b = 4
a > b
So ellipse is along x-axis
c² = a² - b²
c² = 36 - 16 = 20
$c = \sqrt {20} = 2\sqrt 5 $
foci = (±a, 0) =$\left( { \pm 2\sqrt 5 ,0} \right)$
Vertices = (±a, 0) = (±6, 0)
Length of major axis = 2a = 12
Length of miner axis = 2b = 8
eccentricity $ = \frac{c}{a} = \frac{{2\sqrt 5 }}{6} = \frac{{\sqrt 5 }}{3}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 16}}{6} = \frac{{16}}{3}$

Question (2)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 1$

Solution

$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 1$
Comparing to standard form $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ we get
a²= 4 ⇒ a = 2 , a <b
b² = 25 ⇒ b = 5
so ellipse is along y-axis
c² = b² - a² = 25-4 = 21
c = √21
co-ordinate of foci = (0, ±c) = (0, ±√21)
Vertices = (0, ±b) = (0, ±5)
Length of major axis = 2b = 2(5) = 10
Length of miner axis = 2a = 2(2) = 4
eccentricity $e = \frac{c}{b} = \frac{{\sqrt {21} }}{5}$
$LLR = \frac{{2{a^2}}}{b} = \frac{{2 \times 4}}{5} = \frac{8}{5}$

Question (3)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$

Solution

ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$
a² = 16 ⇒ a = 4
b² =a ⇒ b = 3
b>a
ellipse is along x-axis
c² = a²-b² = 16-9 = 7
c = √7
c-ordinates of foci = ( ±c, 0) = (±√7, 0)
vertices = (±a, 0) = (±4, 0)
Length of major axes = 2a = 8
Length of minear axes = 2b = 6
eccentricity $e = \frac{c}{a} = \frac{{\sqrt 7 }}{4}$
Length of latus vectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 9}}{4} = \frac{9}{2}$

Question (4)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{100}} = 1$

Solution

ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{100}} = 1$
a² = 25 ⇒ a = 5
b² = 100 ⇒ b = 10
a < b
ellipse is along y -axis
c² = b² - a² = 100 - 25 = 75
c = √25 = 5√3
co-ordinates of foci = (0, ±c)
co-ordinates of foci =(0, ±5√3)
co-ordinates of vertices =(0, ±b)
co-ordinates of vertices =(0, ±10)
Length of major axis = 2b = 2(10) = 20
Length of minimum axis = 2a = 2(5) = 10
Length of LR $ = \frac{{2{a^2}}}{b} = \frac{{2 \times 25}}{{10}} = 5$
eccentricity $e = \frac{c}{b} = \frac{{5\sqrt 3 }}{{10}} = \frac{{\sqrt 3 }}{2}$

Question (5)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{36}} = 1$

Solution

$\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{36}} = 1$
a² = 49, b² = 36
a = 7 , b= 6, a > b
Ellipse is along x-axis
c² = a² - b² 49-36=13
c = √13
co-ordinate of foci = (±c, 0)
co-ordinate of foci = (±√13, 0)
Co-ordinates of vertices = ( ±a, 0) = (±7, 0)
Length of major axes = 2a = 14
Length of minor axes = 2b = 12
eccentricity $e = \frac{c}{a} = \frac{{\sqrt {13} }}{7}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 36}}{7} = \frac{{72}}{7}$

Question (6)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{400}} = 1$

Solution

$\frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{400}} = 1$
a² = 100 ⇒ a = 10
b² = 400 ⇒ b = 20
a < b
Ellipse is along y-axis
c² = b² - a² = 400 - 100 = 300
c = √300 = 10√
co-ordinates of foci = (0, ±c) = (0, ±10√3)
co-ordinates of vertices = (0, ±b) = (0, ±20)
Length of major axes = 2b = 2(20) = 40
Length of miner axes = 2a = 2(10) = 20
selecticity = $e = \frac{c}{b} = \frac{{10\sqrt 3 }}{{20}} = \frac{{\sqrt 3 }}{2}$
Length of latus rectum $ = \frac{{2{a^2}}}{b} = \frac{{2 \times 100}}{{20}} = 10$

Question (7)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 36x² + 4y² = 144

Solution

36x² + 4y² = 144
$ \Rightarrow \frac{{36{x^2}}}{{144}} + \frac{{4{y^2}}}{{144}} = 1$
$ \Rightarrow \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1$
a² = 4 ⇒ a = 2
b² = 36 ⇒ b = 6
a < b
⇒ Ellipse is along y-axis
c² = b² - a² = 36 - 4 = 32
c= √32 = 4√2
co-ordinate of foci = (0, ±c) = (0, ±4√2)
co-ordinates of vertices = (0, ±b) = (0, ±6)
Length of major axes = 2b = 12
Length of minor axes = 2a = 4
eccentricity $e = \frac{c}{b} = \frac{{4\sqrt 2 }}{6} = \frac{{2\sqrt 2 }}{3}$
Length of latus rectum $ = \frac{{2{a^2}}}{b} = \frac{{2 \times 4}}{6} = \frac{4}{3}$

Question (8)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16x² + y² = 16

Solution

16x² + y² = 16
$\frac{{16{x^2}}}{{16}} + \frac{{{y^2}}}{{16}} = 1$
$\frac{{{x^2}}}{1} + \frac{{{y^2}}}{{16}} = 1$
a² = 1 ⇒ a = 1
b² = 16 ⇒ b = 2
a< b
⇒ Ellipse is along y-axis
c² = b² - a² = 16 - 1 = 15
c= √15
co-ordinate of foci = (0, ±c) = (0, ±√15)
co-ordinates of vertices = (0, ±b) = (0, ±4)
Length of major axes = 2b = 8
Length of minor axes = 2a = 2
eccentricity = $e = \frac{c}{b} = \frac{{\sqrt {15} }}{4}$
Length of latus rectum $ = \frac{{2{a^2}}}{b} = \frac{{2 \times 1}}{4} = \frac{1}{2}$

Question (9)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 4x² + 9y² = 36

Solution

4x² + 9y² = 36
$\frac{{4{x^2}}}{{36}} + \frac{{9{y^2}}}{{36}} = 1$
$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$
a² = 9 ⇒ a = 3
b² = 4 ⇒ b =2
a > b
ellise is along x-axis
c² = a² - b²
c = √5
coordinate of foci = (±c, 0)= (±√5, 0)
coordinate of vertices = (±a, 0) = (±3, 0)
Length of major axis = 2a = 6
Length of minor axis = 2b = 4
eccentricity = $e = \frac{c}{a} = \frac{{\sqrt 5 }}{3}$
Length of latus rectum = $\frac{{2{b^2}}}{a} = \frac{{2 \times 4}}{3} = \frac{8}{3}$

Question (10)

Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci (±4, 0)

Solution

Vertices (±5, 0), foci (±4, 0)
Since y-coordinate of foci is zero
ellipse is alonmg x-axis
∴ vertices = (±a, 0) = (±5, 0) ⇒ a = 5
foci = (±c, 0) = (±4, 0) ⇒ c = 4
∴ c² = a² - b²
∴ 16 = 25 - b²
∴ b² = 9
so equation of ellipse is $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$

Question (11)

Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13), foci (0, ±5)

Solution

Vertices (0, ±13), foci (0, ±5)
As y-coordinate of foci is zero
∴ Ellipse is laong x-axis
vertices = (0, ±b) = (0, ±3) ⇒ b = 13
foci = (0, ±c) = (0, ±5) ⇒ c = 5
∴ c² = a² - b²
25 = 169 - a²
a² = 144 ⇒ a = 12
Equation of ellipse is $\frac{{{x^2}}}{{144}} + \frac{{{y^2}}}{{169}} = 1$

Question (12)

Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci (±4, 0)

Solution

Vertices (±6, 0), foci (±4, 0)
As y-coordinate of foci is zero
∴ Ellipse is laong x-axis
vertices = (±a, 0) = (±6, 0) ⇒ a = 6
foci = (±c, 0) = (±4, 0) ⇒ c = 4
∴ c² = a² - b²
16 = 36 - b²
b² = 20
Equation of ellipse is $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{20}} = 1$

Question (13)

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Solution

Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Length of major axes = 2a = 6, a =3
End points of minor axis = (0, ±b) = (0, ±2), b = 2
Equation of ellipse is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{b^2} = 1$
$=\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$

Question (14)

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $\left( {0, \pm \sqrt 5 } \right)$ , ends of minor axis (±1, 0)

Solution

Ends of major axis = (0, ±a)= $\left( {0, \pm \sqrt 5 } \right)$ , ends of minor axis (±1, 0)
⇒ a = √5, a² = 5
ends of minor axis (±1, 0) = (±1, 0)
⇒ b = 1, b² = 1
Equation of ellipse is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{b^2} = 1$
So equation of ellipse is $\frac{{{x^2}}}{5} + \frac{{{y^2}}}{1} = 1$

Question (15)

Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (±5, 0)

Solution

Length of major axis 26, foci (±5, 0) Since y coordinate of foci is 0, focus is on x-axis, ellipse is along x-axis
∴ length of major axes =2a = 26 ⇒ a = 18
foci = (±c, o) = (±5, 0)
⇒ c = 5
c² = a² - b²
25 = 169 - b²
b² = 169 -25 = 144
So equation of ellipse is $\frac{{{x^2}}}{{169}} + \frac{{{y^2}}}{{144}} = 1$

Question (16)

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ±6)

Solution

Length of minor axis 16, foci (0, ±6)
x coordinate of foci is ero so focus is y-axis, ellipse is along y-axis
length of minor axis = 2a = 16, a =8
foci = (0, ±c) = (0, ±6) ⇒ c = 6
c² = b² - a²
36 = b² - 64
b² = 36+64 = 100
∴ b = 10
So equation of ellipse is $\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{100}} = 1$

Question (17)

Find the equation for the ellipse that satisfies the given conditions: Foci (±3, 0), a = 4

Solution

Foci (±3, 0), a = 4
y coordinate is 0, so ellpse along x-axis a > b
foci (±c, 0) = (±3, 0) ⇒ c = 3
c² = a² - b²
9 = 16 - b²
b² = 7
So equation of ellipse is $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{7} = 1$

Question (18)

Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the x axis.

Solution

b = 3, c = 4, c(0, 0) foci on x-axis
Since foci on x-axis ellipse is along x-axis ∴ a > b
c² = a² - b²
16 = a² - 9
a² = 16 +9 = 25
Equation of ellipse is $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$

Question (19)

Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Solution

c(0, 0) major axes y-axes and passing through (3, 2) and (1, 6)
Major axes is y-axis , so ellipse is along y-axis, so a <b
Let equation be $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
(3, 2) ∈ ellipse
$\therefore \frac{9}{{{a^2}}} + \frac{4}{{{b^2}}} = 1 - - - \left( 1 \right)$
(1, 6) ∈ ellipse
$\therefore \frac{1}{{{a^2}}} + \frac{{36}}{{{b^2}}} = 1 - - - \left( 2 \right)$
Multiply (1) by 9
$\frac{{81}}{{{a^2}}} + \frac{{36}}{{{b^2}}} = 9 - - - \left( 3 \right)$
(3) - (2) ⇒ $\frac{{80}}{{{a^2}}} = 8 \Rightarrow {a^2} = 10$
Replacing value of a² in equation (1)
$\frac{9}{{10}} + \frac{4}{{{b^2}}} = 1$
$\frac{4}{{{b^2}}} = 1 - \frac{9}{{10}} = \frac{1}{{10}}$
b² = 40
So equation of ellipse is $\frac{{{x^2}}}{{10}} + \frac{{{y^2}}}{{40}} = 1$

Question (20)

Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Solution

Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
major axes on x-axis , so ellipse is along x-axis so a>b.
Let equation of ellipse be $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
(4, 3) ∈ ellipse → $\frac{{16}}{{{a^2}}} + \frac{9}{{{b^2}}} = 1 - - - \left( 1 \right)$
(6, 2) ∈ ellipse → $\frac{{36}}{{{a^2}}} + \frac{4}{{{b^2}}} = 1 - - - \left( 2 \right)$
4 ×(1) ⇒ $\frac{{64}}{{{a^2}}} + \frac{{36}}{{{b^2}}} = 4 - - - \left( 3 \right)$
9 × (2) ⇒ $\frac{{324}}{{{a^2}}} + \frac{{36}}{{{b^2}}} = 9 - - - \left( 4 \right)$
(3) - (4)
$\begin{array}{l}\frac{{64}}{{{a^2}}} + \frac{{36}}{{{b^2}}} = 4\\\underline \begin{array}{l}\frac{{324}}{{{a^2}}} + \frac{{36}}{{{b^2}}} = 9\\ - \;\;\quad - \quad \;\; - \end{array} \\ - \frac{{260}}{{{a^2}}}\quad \quad = - 5\end{array}$
${a^2} = \frac{{260}}{5} = 52$
Replace a² in equation (1) we get
$\frac{{16}}{{52}} + \frac{9}{{{b^2}}} = 1$
$\frac{4}{{13}} + \frac{9}{{{b^2}}} = 1$
$\frac{9}{{{b^2}}} = 1 - \frac{4}{{13}} = \frac{9}{{13}}$ b² = 13
So equation of elliipse is $\frac{{{x^2}}}{{52}} + \frac{{{y^2}}}{{13}} = 1$