NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 - Conic Section

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.

$\frac{x^2}{36}+\frac{y^2}{16}=1$

Answer:

Given

The equation of the ellipse

$\frac{x^2}{36}+\frac{y^2}{16}=1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

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

We get

$a=6$ and $b=4$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}$

$c=\sqrt{20}=2\sqrt{5}$

Hence,

Coordinates of the foci:

$(c,0)and(-c,0)=(2\sqrt{5},0)and(-2\sqrt{5},0)$

The vertices:

$(a,0)and(-a,0)=(6,0)and(-6,0)$

The length of the major axis:

$2a=2(6)=12$

The length of minor axis:

$2b=2(4)=8$

The eccentricity :

$e=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(4{)}^2}{6}=\frac{32}{6}=\frac{16}{3}$

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.

$\frac{x^2}{4}+\frac{y^2}{25}=1$

Answer:

Given

The equation of the ellipse

$\frac{x^2}{4}+\frac{y^2}{25}=1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

$a=5$ and $b=2$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}$

$c=\sqrt{21}$

Hence,

Coordinates of the foci:

$(0,c)and(0,-c)=(0,\sqrt{21})and(0,-\sqrt{21})$

The vertices:

$(0,a)and(0,-a)=(0,5)and(0,-5)$

The length of the major axis:

$2a=2(5)=10$

The length of minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{21}}{6}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2{)}^2}{5}=\frac{8}{5}$

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.

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

Answer:

Given

The equation of the ellipse

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

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

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

We get

$a=4$ and $b=3$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}$

$c=\sqrt{7}$

Hence,

Coordinates of the foci:

$(c,0)and(-c,0)=(\sqrt{7},0)and(-\sqrt{7},0)$

The vertices:

$(a,0)and(-a,0)=(4,0)and(-4,0)$

The length of the major axis:

$2a=2(4)=8$

The length of minor axis:

$2b=2(3)=6$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{7}}{4}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(3{)}^2}{4}=\frac{18}{4}=\frac{9}{2}$

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.

$\frac{x^2}{25}+\frac{y^2}{100}=1$

Answer:

Given

The equation of the ellipse

$\frac{x^2}{25}+\frac{y^2}{100}=1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

$a=10$ and $b=5$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{{10}^2-5^2}$

$c=\sqrt{75}=5\sqrt{3}$

Hence,

Coordinates of the foci:

$(0,c)and(0,-c)=(0,5\sqrt{3})and(0,-5\sqrt{3})$

The vertices:

$(0,a)and(0,-a)=(0,10)and(0,-10)$

The length of the major axis:

$2a=2(10)=20$

The length of minor axis:

$2b=2(5)=10$

The eccentricity :

$e=\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(5{)}^2}{10}=\frac{50}{10}=5$

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.

$\frac{x^2}{49}+\frac{y^2}{36}=1$

Answer:

Given

The equation of ellipse

$\frac{x^2}{49}+\frac{y^2}{36}=1\Rightarrow \frac{x^2}{7^2}+\frac{y^2}{6^2}=1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with standard equation of ellipse, which is

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

We get

$a=7$ and $b=6$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}$

$c=\sqrt{13}$

Hence,

Coordinates of the foci:

$(c,0)and(-c,0)=(\sqrt{13},0)and(-\sqrt{13},0)$

The vertices:

$(a,0)and(-a,0)=(7,0)and(-7,0)$

The length of major axis:

$2a=2(7)=14$

The length of minor axis:

$2b=2(6)=12$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{13}}{7}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(6{)}^2}{7}=\frac{72}{7}$

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.

$\frac{x^2}{100}+\frac{y^2}{400}=1$

Answer:

Given

The equation of the ellipse

$\frac{x^2}{100}+\frac{y^2}{400}=1\Rightarrow \frac{x^2}{{10}^2}+\frac{y^2}{{20}^2}=1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

$a=20$ and $b=10$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{{20}^2-{10}^2}$

$c=\sqrt{300}=10\sqrt{3}$

Hence,

Coordinates of the foci:

$(0,c)and(0,-c)=(0,10\sqrt{3})and(0,-10\sqrt{3})$

The vertices:

$(0,a)and(0,-a)=(0,20)and(0,-20)$

The length of the major axis:

$2a=2(20)=40$

The length of minor axis:

$2b=2(10)=20$

The eccentricity :

$e=\frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(10{)}^2}{20}=\frac{200}{20}=10$

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.

$36x^2+4y^2=144$

Answer:

Given

The equation of the ellipse

$36x^2+4y^2=144$

$\Rightarrow \frac{36}{144}{x}^2+\frac{4}{144}{y}^2=1$

$\Rightarrow \frac{1}{4}{x}^2+\frac{1}{36}{y}^2=1$

$\frac{x^2}{2^2}+\frac{y^2}{6^2}=1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

$a=6$ and $b=2$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-2^2}$

$c=\sqrt{32}=4\sqrt{2}$

Hence,

Coordinates of the foci:

$(0,c)and(0,-c)=(0,4\sqrt{2})and(0,-4\sqrt{2})$

The vertices:

$(0,a)and(0,-a)=(0,6)and(0,-6)$

The length of the major axis:

$2a=2(6)=12$

The length of minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2{)}^2}{6}=\frac{8}{6}=\frac{4}{3}$

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.

$16x^2+y^2=16$

Answer:

Given

The equation of the ellipse

$16x^2+y^2=16$

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

$\frac{x^2}{1^2}+\frac{y^2}{4^2}=1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

$a=4$ and $b=1$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-1^2}$

$c=\sqrt{15}$

Hence,

Coordinates of the foci:

$(0,c)and(0,-c)=(0,\sqrt{15})and(0,-\sqrt{15})$

The vertices:

$(0,a)and(0,-a)=(0,4)and(0,-4)$

The length of the major axis:

$2a=2(4)=8$

The length of minor axis:

$2b=2(1)=2$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{15}}{4}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(1{)}^2}{4}=\frac{2}{4}=\frac{1}{2}$

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.

$4x^2+9y^2=36$

Answer:

Given

The equation of the ellipse

$4x^2+9y^2=36$

$\frac{4x^2}{36}+\frac{9y^2}{36}=1$

$\frac{x^2}{9}+\frac{y^2}{4}=1$

$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

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

We get

$a=3$ and $b=2$.

So,

$c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}$

$c=\sqrt{5}$

Hence,

Coordinates of the foci:

$(c,0)and(-c,0)=(\sqrt{5},0)and(-\sqrt{5},0)$

The vertices:

$(a,0)and(-a,0)=(3,0)and(-3,0)$

The length of the major axis:

$2a=2(3)=6$

The length of minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{5}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2{)}^2}{3}=\frac{8}{3}$

Question 10: Find the equation for the ellipse that satisfies the given conditions:

Vertices (± 5, 0), foci (± 4, 0)

Answer:

Given, In an ellipse,

Vertices (± 5, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=5$ and $c=4$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{5^2-4^2}$

$b=\sqrt{9}$

$b=3$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

Which 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)

Answer:

Given, In an ellipse,

Vertices (0, ± 13), foci (0, ± 5)

Here Vertices and focus of the ellipse are in Y-axis so the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=13$ and $c=5$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{{13}^2-5^2}$

$b=\sqrt{169-25}$

$b=\sqrt{144}$

$b=12$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{{12}^2}+\frac{y^2}{{13}^3}=1$

Which 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)

Answer:

Given, In an ellipse,

Vertices (± 6, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=6$ and $c=4$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{6^2-4^2}$

$b=\sqrt{36-16}$

$b=\sqrt{20}$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{6^2}+\frac{y^2}{(\sqrt{20}{)}^2}=1$

Which 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)

Answer:

Given, In an ellipse,

Ends of the major axis (± 3, 0), ends of minor axis (0, ± 2)

Here, the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get

$a=3$ and $b=2$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$

Which is

$\frac{x^2}{9}+\frac{y^2}{4}=1$.

Question 14: Find the equation for the ellipse that satisfies the given conditions:

Ends of major axis (0, ± $\sqrt{5}$ ), ends of minor axis (± 1, 0)

Answer:

Given, In an ellipse,

Ends of the major axis (0, ±$\sqrt{5}$ ), ends of minor axis (± 1, 0)

Here, the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get

$a=\sqrt{5}$ and $b=1$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{5}{)}^2}=1$

Which is

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

Question 15: Find the equation for the ellipse that satisfies the given conditions:

Length of major axis 26, foci (± 5, 0)

Answer:

Given, In an ellipse,

Length of major axis 26, foci (± 5, 0)

Here, the focus of the ellipse is in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( Length of semimajor axis and foci) with the given one, we get

$2a=26\Rightarrow a=13$ and $c=5$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{{13}^2-5^2}$

$b=\sqrt{144}$

$b=12$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{{13}^2}+\frac{y^2}{{12}^2}=1$

Which 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).

Answer:

Given, In an ellipse,

Length of minor axis 16, foci (0, ± 6).

Here, the focus of the ellipse is on the Y-axis so the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( length of semi-minor axis and foci) with the given one, we get

$2b=16\Rightarrow b=8$ and $c=6$

Now, As we know the relation,

$a^2=b^2+c^2$

$a=\sqrt{b^2+c^2}$

$a=\sqrt{8^2+6^2}$

$a=\sqrt{64+36}$

$a=\sqrt{100}$

$a=10$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{8^2}+\frac{y^2}{{10}^3}=1$

Which 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

Answer:

Given, In an ellipse,

V Foci (± 3, 0), a = 4

Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=4$ and $c=3$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{4^2-3^2}$

$b=\sqrt{7}$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{4^2}+\frac{y^2}{(\sqrt{7}{)}^2}=1$

Which 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.

Answer:

Given,In an ellipse,

b = 3, c = 4, centre at the origin; foci on the x axis.

Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

Also Given,

$b=3$ and $c=4$

Now, As we know the relation,

$a^2=b^2+c^2$

$a^2=3^2+4^2$

$a^2=25$

$a=5$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

Which 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).

Answer:

Given,in an ellipse

Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Since, The major axis of this ellipse is on the Y-axis, the equation of the ellipse will be of the form:

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

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

Now since the ellipse passes through points,(3, 2)

$\frac{3^2}{b^2}+\frac{2^2}{a^2}=1$

$9a^2+4b^2=a^2{b}^2$

since the ellipse also passes through points,(1, 6).

$\frac{1^2}{b^2}+\frac{6^2}{a^2}=1$

$a^2+36b^2=a^2{b}^2$

On solving these two equation we get

$a^2=40$ and $b^2=10$

Thus, The equation of the ellipse will be

$\frac{x^2}{10}+\frac{y^2}{40}=1$.

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).

Answer:

Given, in an ellipse

Major axis on the x-axis and passes through the points (4,3) and (6,2).

Since The major axis of this ellipse is on the X-axis, the equation of the ellipse will be of the form:

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

Where $a$ and $b$are the length of the semimajor axis and semiminor axis respectively.

Now since the ellipse passes through the point,(4,3)

$\frac{4^2}{a^2}+\frac{3^2}{b^2}=1$

$16b^2+9a^2=a^2{b}^2$

since the ellipse also passes through the point (6,2).

$\frac{6^2}{a^2}+\frac{2^2}{b^2}=1$

$4a^2+36b^2=a^2{b}^2$

On solving this two equation we get

$a^2=52$ and $b^2=13$

Thus, The equation of the ellipse will be

$\frac{x^2}{52}+\frac{y^2}{13}=1$