Class 11 Maths NCERT Solutions Chapter 10 Ex 10.3 – Conic Sections

Since 36>16, the major axis is along the x-axis. Comparing with \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1: a=6,\ b=4.

c=\sqrt{a^2-b^2}=\sqrt{36-16}=\sqrt{20}=2\sqrt5

Since 25>4, the major axis is along the y-axis. Comparing with \dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1: a=5,\ b=2.

c=\sqrt{a^2-b^2}=\sqrt{25-4}=\sqrt{21}

Since 16>9, the major axis is along the x-axis. Here a=4,\ b=3.

c=\sqrt{16-9}=\sqrt7

Since 100>25, the major axis is along the y-axis. Here a=10,\ b=5.

c=\sqrt{100-25}=\sqrt{75}=5\sqrt3

Since 49>36, the major axis is along the x-axis. Here a=7,\ b=6.

c=\sqrt{49-36}=\sqrt{13}

Since 400>100, the major axis is along the y-axis. Here a=20,\ b=10.

c=\sqrt{400-100}=\sqrt{300}=10\sqrt3

Dividing throughout by 144 to reach standard form:

\dfrac{x^2}{4}+\dfrac{y^2}{36}=1

Since 36>4, the major axis is along the y-axis. Here a=6,\ b=2.

c=\sqrt{36-4}=\sqrt{32}=4\sqrt2

Dividing throughout by 16 to reach standard form:

\dfrac{x^2}{1}+\dfrac{y^2}{16}=1

Since 16>1, the major axis is along the y-axis. Here a=4,\ b=1.

c=\sqrt{16-1}=\sqrt{15}

Dividing throughout by 36 to reach standard form:

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

Since 9>4, the major axis is along the x-axis. Here a=3,\ b=2.

c=\sqrt{9-4}=\sqrt5

Since the vertices and foci lie on the x-axis: a=5,\ c=4.

b^2=a^2-c^2=25-16=9

Since the vertices and foci lie on the y-axis: a=13,\ c=5.

b^2=a^2-c^2=169-25=144

Here a=6,\ c=4.

b^2=a^2-c^2=36-16=20

The major axis lies on the x-axis, giving a=3; the minor axis gives b=2.

The major axis lies on the y-axis, giving a=\sqrt5; the minor axis gives b=1.

The major axis has length 2a=26 \ \Rightarrow\ a=13. The foci give c=5.

b^2=a^2-c^2=169-25=144

The minor axis has length 2b=16 \ \Rightarrow\ b=8. The foci lie on the y-axis, giving c=6.

a^2=b^2+c^2=64+36=100

Here c=3,\ a=4.

b^2=a^2-c^2=16-9=7

Using a^2=b^2+c^2:

a^2=9+16=25

With the major axis on the y-axis, the equation has the form \dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1.

Substituting (3,2):

\dfrac{9}{b^2}+\dfrac{4}{a^2}=1 \qquad\ldots(1)

Substituting (1,6):

\dfrac{1}{b^2}+\dfrac{36}{a^2}=1 \qquad\ldots(2)

Let u=\dfrac{1}{b^2} and v=\dfrac{1}{a^2}. Then (1) and (2) become:

9u+4v=1 \qquad\ldots(1')

u+36v=1 \qquad\ldots(2')

From (2'): u=1-36v. Substituting into (1'):

9(1-36v)+4v=1 \ \Rightarrow\ 9-324v+4v=1 \ \Rightarrow\ -320v=-8 \ \Rightarrow\ v=\dfrac{1}{40}

Then u=1-36\left(\dfrac{1}{40}\right)=1-\dfrac{9}{10}=\dfrac{1}{10}. So a^2=\dfrac{1}{v}=40 and b^2=\dfrac{1}{u}=10.

With the major axis on the x-axis, the equation has the form \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.

Substituting (4,3):

\dfrac{16}{a^2}+\dfrac{9}{b^2}=1 \qquad\ldots(1)

Substituting (6,2):

\dfrac{36}{a^2}+\dfrac{4}{b^2}=1 \qquad\ldots(2)

Let u=\dfrac{1}{a^2} and v=\dfrac{1}{b^2}. Then (1) and (2) become:

16u+9v=1 \qquad\ldots(1')

36u+4v=1 \qquad\ldots(2')

Multiplying (1') by 4 and (2') by 9, then subtracting to eliminate v:

64u+36v=4 \qquad 324u+36v=9

260u=5 \ \Rightarrow\ u=\dfrac{1}{52}

Substituting back into (1'): 16\left(\dfrac{1}{52}\right)+9v=1 \ \Rightarrow\ \dfrac{4}{13}+9v=1 \ \Rightarrow\ 9v=\dfrac{9}{13} \ \Rightarrow\ v=\dfrac{1}{13}.

So a^2=\dfrac{1}{u}=52 and b^2=\dfrac{1}{v}=13.