Class 11 Maths NCERT Solutions Chapter 10 Ex 10.4 – Conic Sections

Comparing with \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1: a=4,\ b=3.

c=\sqrt{a^2+b^2}=\sqrt{16+9}=\sqrt{25}=5

Since y^2 is the positive term, the transverse axis is along the y-axis. Comparing with \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1: a=3,\ b=\sqrt{27}=3\sqrt3.

c=\sqrt{a^2+b^2}=\sqrt{9+27}=\sqrt{36}=6

Dividing throughout by 36 to reach standard form:

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

The transverse axis is along the y-axis, with a=2,\ b=3.

c=\sqrt{4+9}=\sqrt{13}

Dividing throughout by 576 to reach standard form:

\dfrac{x^2}{36}-\dfrac{y^2}{64}=1

The transverse axis is along the x-axis, with a=6,\ b=8.

c=\sqrt{36+64}=\sqrt{100}=10

Dividing throughout by 36 to reach standard form:

\dfrac{y^2}{36/5}-\dfrac{x^2}{4}=1

The transverse axis is along the y-axis, with a^2=\dfrac{36}{5},\ b^2=4, so a=\dfrac{6}{\sqrt5}=\dfrac{6\sqrt5}{5}.

c^2=a^2+b^2=\dfrac{36}{5}+4=\dfrac{56}{5} \ \Rightarrow\ c=\sqrt{\dfrac{56}{5}}=\dfrac{2\sqrt{70}}{5}

e=\dfrac{c}{a}=\sqrt{\dfrac{56/5}{36/5}}=\sqrt{\dfrac{56}{36}}=\dfrac{\sqrt{14}}{3}

Dividing throughout by 784 to reach standard form:

\dfrac{y^2}{16}-\dfrac{x^2}{49}=1

The transverse axis is along the y-axis, with a=4,\ b=7.

c=\sqrt{16+49}=\sqrt{65}

Here a=2,\ c=3.

b^2=c^2-a^2=9-4=5

Since the vertices and foci lie on the y-axis: a=5,\ c=8.

b^2=c^2-a^2=64-25=39

Here a=3,\ c=5.

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

The transverse axis has length 2a=8 \ \Rightarrow\ a=4. The foci give c=5.

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

The conjugate axis has length 2b=24 \ \Rightarrow\ b=12. The foci lie on the y-axis, giving c=13.

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

Here c=3\sqrt5, so c^2=45. The latus rectum gives \dfrac{2b^2}{a}=8 \ \Rightarrow\ b^2=4a.

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

45=a^2+4a \ \Rightarrow\ a^2+4a-45=0

Factoring: (a-5)(a+9)=0 \ \Rightarrow\ a=5 (rejecting the negative root).

Then b^2=4(5)=20.

Here c=4, so c^2=16. The latus rectum gives \dfrac{2b^2}{a}=12 \ \Rightarrow\ b^2=6a.

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

16=a^2+6a \ \Rightarrow\ a^2+6a-16=0

Factoring: (a-2)(a+8)=0 \ \Rightarrow\ a=2 (rejecting the negative root).

Then b^2=6(2)=12.

Here a=7. Since e=\dfrac{c}{a}:

c=ae=7\times\dfrac43=\dfrac{28}{3}

b^2=c^2-a^2=\left(\dfrac{28}{3}\right)^2-49=\dfrac{784}{9}-\dfrac{441}{9}=\dfrac{343}{9}

Since the foci lie on the y-axis, the equation has the form \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1, with c^2=a^2+b^2=10, i.e. b^2=10-a^2.

Substituting the point (2,3):

\dfrac{9}{a^2}-\dfrac{4}{10-a^2}=1

Multiplying through by a^2(10-a^2):

9(10-a^2)-4a^2=a^2(10-a^2)

90-9a^2-4a^2=10a^2-a^4 \ \Rightarrow\ a^4-23a^2+90=0

This is a quadratic in a^2. Solving:

a^2=\dfrac{23\pm\sqrt{529-360}}{2}=\dfrac{23\pm13}{2} \ \Rightarrow\ a^2=18 \text{ or } a^2=5

If a^2=18, then b^2=10-18=-8, which is not valid since b^2 must be positive. So a^2=5, giving b^2=10-5=5.