Exercise 5.2-Additional Problems - Chapter 5 - Two Dimensional Analytical Geometry – II - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Find the equation of the ellipse if centre is $(3,-4)$, one of the foci is $(3+\sqrt{3},-4)$ and $e=\frac{\sqrt{3}}{2}$ Solution:
Given Centre $=\mathrm{C}=(3,-4)$ and $\mathrm{F}=(3+\sqrt{3},-4)$ and $e=\frac{\sqrt{3}}{2}$
$
\begin{array}{cc}
\Rightarrow & \mathrm{CF}=a e=\sqrt{(3+\sqrt{3}-3)^2+(-4+4)^2}=\sqrt{3} \\
\Rightarrow & a\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \Rightarrow a=2 \\
b^2=a^2\left(1-e^2\right)=4\left(1-\frac{3}{4}\right)=1
\end{array}
$
Major axis is along a line parallel to $\mathrm{X}$ - axis.
$\therefore$ Equation of the ellipse is $\frac{(x-3)^2}{4}+\frac{(y+4)^2}{1}=1$.
Question 2.
Find the equation of the hyperbola if centre $(1,-2)$; length of the transverse axis is $8 ; \mathrm{e}=\frac{5}{4}$ and the transverse axis is parallel to $\mathrm{X}$-axis.
Solution:
Here, $2 a=8 \quad \Rightarrow a=4$ and $a^2=16$
$
\begin{aligned}
e & =\frac{5}{4} \\
b^2 & =a^2\left(e^2-1\right)=16\left(\frac{25}{16}-1\right)=16\left(\frac{9}{16}\right)=9
\end{aligned}
$
Here, centre $=(1,-2)$ and transverse axis is parallel to $\mathrm{X}$-axis.
$\therefore$ Equation of the hyperbola is $\frac{(x-1)^2}{16} \quad \frac{(y+2)^2}{9}=1$.
Question 3.
Find axis, vertex, focus and equation of directrix for $y^2+8 x-6 y+1=0$

Solution:

$\begin{aligned}
& y^2-6 y=-8 x-1 \\
& y^2-6 y+9=-8 x-1+9 \\
& (y-3)^2=-8 x+8=-8(x-1)
\end{aligned}$

Comparing this equation with $\mathrm{Y}^2=-4 \mathrm{aX}$ we get $4 \mathrm{a}=8$ or $\mathrm{a}=2$
Vertex is $(0,0)$ $\mathrm{x}-1=0 \Rightarrow \mathrm{x}=1, \mathrm{y}-3=0 \Rightarrow \mathrm{y}=3$
Axis $y-3=0$, Vertex $=(1,3)$
Focus is $(-\mathrm{a}, 0)=(-2,0)+(1,3)=(-1,3)$
Equation of directrix is $x-a=0$. i.e., $X-2=0$
$\Rightarrow \mathrm{x}-1-2=0 \Rightarrow \mathrm{x}-3=0$
Latus rectum $x+a=0$ i.e., $x-1+2=0$ $\mathrm{x}+1=0$
Length of latus rectum $=4 a=8$
Question 4.
Find axis, Vertex focus and equation of directrix for $x^2-6 x-12 y-3=0$.
Solution:
$
\begin{aligned}
& x^2-6 x=12 y+3 \\
& x^2-6 x+9=12 y+3+9=12 y+12 \\
& (x-3)^2=12(y+1)
\end{aligned}
$

This equation is of the form $X^2=4 a Y$
$
\begin{aligned}
& 4 a=12 \\
& \Rightarrow a=3
\end{aligned}
$
Vertex is $\mathrm{x}-3=0 ; \mathrm{y}+1=0$
$
\Rightarrow \mathrm{x}=3 ; \mathrm{y}=-1
$

Question 5.
Find the eccentricity, centre, foci and vertices of the following hyperbolas: $x^2-4 y^2-8 x-6 y-18=0$

Solution:
$
\begin{aligned}
x^2+4 y^2-8 x-6 y-68 & =0 \\
\frac{(x-4)^2}{100}+\frac{(y-2)^2}{25} & =1 \\
\text { Taking } x-4=\mathrm{X}, y-2 & =\mathrm{Y} \text { we get, } \\
\frac{x^2}{100}+\frac{y^2}{25} & =1 \\
\text { Major axis } & =\mathrm{X} \text {-axis } \\
a^2 & =100, b^2=25 \\
e & =\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{25}{100}}=\sqrt{\frac{75}{100}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \\
\text { Centre } & =(4,2) \\
a & =10, e=\frac{\sqrt{3}}{2} \Rightarrow a e=\frac{10 \sqrt{3}}{2}=5 \sqrt{3}
\end{aligned}
$

Question 6.
Find the eccentricity, centre, foci, vertices of $9 x^2+4 y^2=36$
Solution:
$
\begin{aligned}
(\div \text { by 36 }) \Rightarrow \frac{9 x^2}{36}+\frac{4 y^2}{36} & =1 \\
\frac{x^2}{4}+\frac{y^2}{9} & =1 \\
\mathrm{Y}-\mathrm{axis} & =\text { Major axis, } \\
a^2 & =9 \Rightarrow a=3 \\
b^2 & =4 \Rightarrow b=2 \\
e & =\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3} \\
\text { Centre } & =(0,0) \\
a & =3, e=\frac{\sqrt{5}}{3}, \therefore a e=\sqrt{5} \\
\text { Foci }=(0, \pm a e) & =(0, \pm \sqrt{5}) ; \text { Vertices }=(0, \pm a)=(0, \pm 3)
\end{aligned}
$

Question 7.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
$
x^2-4 y^2+6 x+16 y-11=0
$
Solution:
$
\begin{aligned}
x^2-4 y^2+6 x+16 y-11 & =0 \\
\left(x^2+6 x\right)-\left(4 y^2-16 y\right) & =11 \\
\left(x^2+6 x+9-9\right)-4\left(y^2-4 y\right) & =11 \\
(x+3)^2-9-4\left(y^2-4 y+4-4\right) & =11 \\
(x+3)^2-9-4(y-2)^2+16 & =11 \\
(x+3)^2-4(y-2)^2 & =11-16+9=4 \\
(\div \text { by } 4) \Rightarrow \frac{(x+3)^2}{4}-\frac{(y-2)^2}{1} & =1
\end{aligned}
$

$\begin{aligned}
a^2 & =4, b^2=1 \\
e & =\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{4}}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2} \\
\text { Centre } & =(-3,2) \\
a & =2, e=\frac{\sqrt{5}}{2}, a e=2 \times \frac{\sqrt{5}}{2}=\sqrt{5} \\
\text { Foci }=(\pm a e, 0) & =(\pm \sqrt{5}, 0) \\
\mathrm{F}_1=(\sqrt{5}, 0)+(-3,2) & =(-3+\sqrt{5}, 2) \\
\mathrm{F}_2=(-\sqrt{5}, 0)+(-3,2) & =[-3-\sqrt{5}, 2] \\
\text { Vertices }=(\pm a, 0) & =(\pm 2,0) \\
\mathrm{V}_1 & =(2,0)+(-3,2)=(-1,2) \\
\mathrm{V}_2 & =(-2,0)+(-3,2)=(-5,2)
\end{aligned}$

Question 8.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
$
x^2-3 y^2+6 x+6 y+18=0
$
Solution:
$
\begin{aligned}
& x^2-3 y^2+6 x+6 y+18=0 \\
& \left(x^2+6 x\right)-\left(3 y^2-6 y\right)=-18 \\
& \left(x^2+6 x+9-9\right)-3\left(y^2-2 y+1-1\right)=-18 \\
& (x+3)^2-9-3(y-1)^2+3=-18 \\
& (x+3)^2-3(y-1)^2=-18+9-3=-12 \\
& \text { So, } \quad 3(y-1)^2-(x+3)^2=12 \\
& (\div \text { by } 12) \Rightarrow \frac{(y-1)^2}{4}-\frac{(x+3)^2}{12}=1 \\
& \text { Transverse axis is parallel to } \mathrm{Y} \text {-axis. } \\
& a^2=4, b^2=12 \\
& e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{12}{4}}=\sqrt{1+3}=2 \\
& a=2, e=2, a e=2 \times 2=4 ; e=2 \\
& \text { Centre }=(-3,1) ; \quad \text { Foci }=(0, \pm a e)=(0, \pm 4) \\
& \mathrm{F}_1=(0,4)+(-3,1)=(-3,5) \\
& \mathrm{F}_2=(0,-4)+(-3,1)=(-3,-3) \\
& \text { Vertices }(0, \pm a)=(0, \pm 2) \\
& \mathrm{V}_1=(0,2)+(-3,1)=(-3,3) \\
& \mathrm{V}_2=(0,-2)+(-3,1)=(-3,-1) \\
&
\end{aligned}
$