Exercise 5.2 - Chapter 5 Two Dimensional Analytical Geometry–II 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $5.2$
Question 1.
Find the equation of the parabola in each of the cases given below:
(i) focus $(4,0)$ and directrix $x=-4$.
(ii) passes through $(2,-3)$ and symmetric about $y$-axis.
(iii) vertex $(1,-2)$ and focus $(4,-2)$.
(iv) end points of latus rectum $(4,-8)$ and $(4,8)$.
Solution:
(i) Focus $=\mathrm{F}=(4,0)$
$\Rightarrow \mathrm{a}=4$
Equation of directrix $\mathrm{x}=-4$
$\Rightarrow$ The curve open to the right. So the equation will be of the form $y^2=4 \mathrm{ax}$ Here $\mathrm{a}=4$
$
\Rightarrow y^2=4 \text { (4) } x \text { (i.e.) } y^2=16 x
$

(ii) The parabola is symmetric about $\mathrm{y}$ axis. So the equation will be of the form $x^2=4$ ay
It passes through $(2,-3)$
$
\begin{aligned}
& \Rightarrow 2^2=4 \mathrm{a}(-3) \\
& 4=-12 \mathrm{a} \Rightarrow \mathrm{a}=-\frac{1}{3} \Rightarrow 4 \mathrm{a}=-\frac{4}{3} \\
& \therefore \text { Equation of parabola is } \mathrm{x}^2=-\frac{4}{3} \mathrm{y} \\
& 3 \mathrm{x}^2=-4 \mathrm{y} \text {. }
\end{aligned}
$
(iii) The distance between vertex and focus $=3$
(ie.,) $\mathrm{a}=3$
Parabola is open to the right.
So equation will be of the form $y^2=4 a x$
Here $\mathrm{a}=3 \Rightarrow \mathrm{y}^2=12 \mathrm{x}$
but the vertex is $(1,-2)$

So equation of the parabola is
$
(\mathrm{y}+2)^2=12(\mathrm{x}-1)
$

(iv) Focus $=(4,0)$
Equation of the parabola will be of the form . $y^2=4 a x$
Here $\mathrm{a}=4$
$
\Rightarrow y^2=16 x
$

Question 2.
Find the equation of the ellipse in each of the cases given below:
(i) foci $(\pm 3,0), \mathrm{e}=\frac{1}{2}$
(ii) foci $(0, \pm 4)$ and end points of major axis are $(0, \pm 5)$.
(iii) length of latus rectum 8 , eccentricity $=\frac{3}{5}$ and major axis on $x$-axis.
(iv) length of latus rectum 4 , distance between foci $4 \sqrt{2}$ and major axis as $y$-axis.
Solution:
(i) Given ae $=3$ and e $=\frac{1}{2}$
$\Rightarrow \mathrm{a}\left(\frac{1}{2}\right)=3 \Rightarrow \mathrm{a}=6$
So $\mathrm{a}^2=36$
$
\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=36\left(1-\frac{1}{4}\right)=36 \times \frac{3}{4}=27
$
Since Foci $=(\pm 3,0)$, major axis is along $x$-axis
So equation of ellipse is $\frac{x^2}{36}+\frac{y^2}{27}=1$

(ii) From the diagram we see that major axis is along $y$-axis.
Also $\mathrm{a}=5$ and ae $=4$

$
\Rightarrow 5 \mathrm{e}=4 \Rightarrow \mathrm{e}=\frac{4}{5}
$
Now $a=5 \Rightarrow a^2=25$
$
\text { ae }=4 \Rightarrow a^2=16
$
We know $b^2=a^2\left(1-e^2\right)=a^2-a^2 e^2=25-16=9$
Equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{25}=1$

(iii) Given $\frac{2 b^2}{a}=8$ and $e=\frac{3}{5}$
Now
But $\quad b^2=a^2\left(1-e^2\right) \quad \Rightarrow 4 a=a^2\left(1-\left(\frac{3}{5}\right)^2\right)=a^2\left(1-\frac{9}{25}\right)$
$
\frac{2 b^2}{a}=8 \quad \Rightarrow b^2=\frac{8 a}{2}=4 a
$
$\Rightarrow \quad 4 a=\frac{16 a^2}{25} \quad \Rightarrow a=\frac{100}{16}=\frac{25}{4}$
So
$a^2=\frac{625}{16}$ and $b^2=4 a=\frac{625}{16}=4 \times \frac{25}{4}=25$
Here major axis is along $x$-axis
Equation of ellipse is
$
\frac{x^2}{\frac{625}{16}}+\frac{y^2}{25}=1 \quad \Rightarrow \frac{16 x^2}{625}+\frac{y^2}{25}=1
$

(iv) Given $\frac{2 b^2}{a}=4$ and $2 \mathrm{ae}=4 \sqrt{2}$
Now $\frac{2 b^2}{a}=42 b^2=4 \mathrm{a}$
$
\begin{aligned}
& \Rightarrow \mathrm{b}^2=2 \mathrm{a} \\
& 2 \mathrm{ae}=4 \sqrt{2} \mathrm{ae}=2 \sqrt{2}
\end{aligned}
$
So $\mathrm{a}^2 \mathrm{e}^2=4(2)=8$
We know $b^2=a^2\left(1-e^2\right)=a^2-a^2 e^2$
$
\begin{aligned}
& \Rightarrow 2 a=a^2-8 \Rightarrow a^2-2 a-8=0 \\
& \Rightarrow(a-4)(a+2)=0 \Rightarrow a=4 \text { or }-2
\end{aligned}
$
As a cannot be negative
$
a=4 \text { So } a^2=16 \text { and } b^2=2(4)=8
$
Also major axis is along $j$-axis
So equation of ellipse is $\frac{x^2}{8}+\frac{y^2}{16}=1$

Question 3.
Find the equation of the hyperbola in each of the cases given below:
(i) foci $(\pm 2,0)$, eccentricity $=\frac{3}{2}$
(ii) Centre $(2,1)$, one of the foci $(8,1)$ and corresponding directrix $x=4$.
(iii) passing through $(5,-2)$ and length of the transverse axis along JC axis and of length 8 units.
Solution:
(i) Given
$
\begin{aligned}
& \mathrm{ae}=2 \text { and } \mathrm{e}=\frac{3}{2} \\
& \mathrm{a}\left(\frac{3}{2}\right)=2 \Rightarrow \mathrm{a}=\frac{4}{3} \text { So } \mathrm{a}^2=\frac{16}{9} \\
& \mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)=\mathrm{a}^2 \mathrm{e}^2-\mathrm{a}^2=4-\frac{16}{9}=\frac{20}{9}
\end{aligned}
$
Since the foci are $(\pm 2,0)$, transverse axis is along $\mathrm{x}$-axis
So equation of hyperbola is
$
\frac{x^2}{16 / 9}-\frac{y^2}{20 / 9}=1 \Rightarrow \frac{9 x^2}{16}-\frac{9 y^2}{20}=1
$
(ii) Given Centre $=(2,1)$
ae $=6$ (distance between $(2,1)$ and $(8,1))$
Also $\frac{a}{e}=2 \Rightarrow \mathrm{a}=2 \mathrm{e}$
Equation of directrix is $x=4$ [(i.e.,) $(x-2=2)$ Since centre is $(2,1)$ ] $\Rightarrow \frac{a}{e}=2$
Given ae $=6 \Rightarrow \mathrm{a}^2 \mathrm{e}^2=36$
$
\begin{aligned}
& \text { (i.e.) }(2 \mathrm{e})^2(\mathrm{e})^2=36 \\
& \Rightarrow 4 \mathrm{e}^4=36 \Rightarrow \mathrm{e}^4=9 \\
& \Rightarrow \mathrm{e}=\sqrt{3}
\end{aligned}
$
Now e $=\sqrt{3} \mathrm{a}=2 \sqrt{3}$
$
\begin{aligned}
& \therefore \mathrm{a}^2=4 \times 3=12 \\
& \mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)=\mathrm{a}^2 \mathrm{e}^2-\mathrm{a}^2=36-12=24
\end{aligned}
$
So here Centre $=(2,1)$
So equation of hyperbola is
$
\frac{(x-2)^2}{12}-\frac{(y-1)^2}{24}=1
$
(iii) Length of the transverse axis $=8$
$
2 \mathrm{a}=8 \Rightarrow \mathrm{a}=4
$
Transverse axis is along $\mathrm{x}$-axis

So of equation of hyperbola is will be
$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad \text { (ie.,) } \frac{x^2}{16}-\frac{y^2}{b^2}=1
$
But the hyperbola passes through $(5,-2)$
$
\begin{aligned}
& \Rightarrow \quad \frac{25}{16}-\frac{4}{b^2}=1 \Rightarrow \frac{25}{16}=1+\frac{4}{b^2} \\
& \text { (or) } \frac{25}{16}-1=\frac{4}{b^2} \Rightarrow \frac{4}{b^2}=\frac{9}{16} \\
& 9 b^2=64 \quad \Rightarrow b^2=\frac{64}{9} \\
&
\end{aligned}
$
$
\frac{x^2}{16}-\frac{y^2}{649}=1 \Rightarrow \frac{x^2}{16}-\frac{9 y^2}{64}=1
$

Question 4.
Find the vertex, focus, equation of directrix and length of the latus rectum of the following:
(i) $y^2=16 x$
(ii) $\mathrm{x}^2=24 \mathrm{y}$
(iii) $y^2=-8 x$
(iv) $x^2-2 x+8 y+17=0$
(v) $y^2-4 y-8 x+12=0$
Solution:
(i) $y^2=16 x$
It is of the form $y^2=4 \mathrm{ax}$ (type I)
Here $4 a=16 \Rightarrow a=4$
Vertex $=(0,0)$
Focus $=(a, 0)=(4,0)$
Equation of directrix $x+4=0$ (or) $x=-4$
Length of latus rectum $=4 a=16$.
(ii) $x^2=24 y$
This is of the form $\mathrm{x}^2=4$ ay (type III) $4 a=24 \Rightarrow a=6$
Vertex $=(0,0)$
Focus $=(0, a)=(0,6)$
Equation of directrix is $y+a=0$ (i.e.,) $y+6=0$ (or) $y=-6$
Length of latus rectum $=4 a=24$.
(iii) $y^2=-8 x$
This is of the form $y^2=-4 a x$ (type II)
Here $4 a=8 \Rightarrow a=2$
Vertex $=(0,0)$
Focus $=(-\mathrm{a}, 0)=(-2,0)$
Equation of directrix is $\mathrm{x}-2=0$ (or) $\mathrm{x}=2$
Length of latus rectum $=4 a=8$.

$
\begin{aligned}
& \text { (iv) } x^2-2 x+8 y+17=0 \\
& x^2-2 x=-8 y-17 \\
& x^2-2 x+1-1=-8 y-17 \\
& (x-1)^2=-8 y-17+1=-8 y+16 \\
& (x-1)^2=-8(y-2)
\end{aligned}
$
Taking $\mathrm{x}-1=\mathrm{X}$ and $\mathrm{y}-2=\mathrm{Y}$.
We get $X^2=-8 Y$.
This is of the form $x^2=-4$ ay (type IV)
Where $4 a=8 \Rightarrow a=2$

$
\begin{aligned}
& \text { (v) } y^2-4 y=8 x-12=0 \\
& y^2-4 y+4=8 x-12+4 \\
& (y-2)^2=8 x-8=8(x-1)
\end{aligned}
$
Taking $\mathrm{x}-1=\mathrm{X}$ and $\mathrm{y}-2=\mathrm{Y}$.
We get $Y^2=8 \mathrm{X}$.
This is of the form $y^2=4 a x$ (type IV)
Where $4 a=8 \Rightarrow a=2$

Question 5.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:
(i) $\frac{x^2}{25}+\frac{y^2}{9}=1$
(ii) $\frac{x^2}{3}+\frac{y^2}{10}=1$
(iii) $\frac{x^2}{25}-\frac{y^2}{144}=1$
(iv) $\frac{y^2}{16}-\frac{x^2}{9}=1$
Solution:
(i) $\frac{x^2}{25}+\frac{y^2}{9}=1$
It is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, which is an ellipse
Here $\mathrm{a}^2=25, \mathrm{~b}^2=9$
$\mathrm{a}=5, \mathrm{~b}=3$
$\mathrm{e}^2=\frac{a^2-b^2}{a^2}=\frac{25-9}{25}=\frac{16}{25} \Rightarrow \mathrm{e}=\frac{4}{5}$
Now $\mathrm{e}=\frac{4}{5}$ and $\mathrm{a}=5 \Rightarrow \mathrm{ae}=4$ and $\frac{a}{e}=\frac{5}{4 / 5}=\frac{25}{4}$
Here the major axis is along $\mathrm{x}$ axis
$\therefore$ Centre $=(0,0)$
Foci $=(\pm a e, 0)=(\pm 4,0)$
Vertices $=(\pm \mathrm{a}, 0)=(\pm 5,0)$
Equation of directrix $\mathrm{x}=\pm \frac{a}{e}$ (ie.,) $\mathrm{x}=\pm \frac{25}{4}$

(ii) $\frac{x^2}{3}+\frac{y^2}{10}=1$
It is an ellipse and here (always $a>b$ )
Here
$
\begin{aligned}
a^2 & =10, b^2=3 \\
a & =\sqrt{10} \\
e^2 & =\frac{a^2-b^2}{a^2}=\frac{10-3}{10}=\frac{7}{10} \quad \Rightarrow e=\frac{\sqrt{7}}{\sqrt{10}}
\end{aligned}
$
Now
Here the major axis is alongy-axis
Centre $=(0,0)$
Foci $=(0, \pm \mathrm{ae})=(0, \pm \sqrt{7})$
Vertices $=(0, \pm a)=(0, \pm \sqrt{10})$
Equation of directrices $y=\pm \frac{10}{\sqrt{7}}$
(iii) $\frac{x^2}{25}-\frac{y^2}{144}=1$
This is a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
Here
$
\begin{aligned}
a^2 & =25, b^2=144 \\
a & =5 \\
e^2 & =\frac{a^2+b^2}{a^2}=\frac{25+144}{25}=\frac{169}{25} \quad \Rightarrow e=\frac{13}{5}
\end{aligned}
$
Now $e=\frac{13}{5}$ and $a=5 \quad \Rightarrow a e=13$ and $\frac{a}{e}=\frac{5}{13 / 5}=\frac{25}{13}$
Now, Here transverse axis is along $\mathrm{x}$-axis
Centre $=(0,0)$
Vertices $=(\pm \mathrm{a}, 0)=(\pm 5,0)$
Foci $=(\pm$ ae, 0$)=(\pm 13,0)$
Equation of directrices $x=\pm \frac{a}{e}$ (ie.,) $x=\pm \frac{25}{13}$(iv) $\frac{y^2}{16}-\frac{x^2}{9}=1$
It is a hyperbola. Here transverse axis is along y-axis ..
Also $a^2=16$ and $b^2=9$
Now
$
e^2=\frac{a^2+b^2}{a^2}=\frac{16+9}{16}=\frac{25}{16} \quad \Rightarrow e=\frac{5}{4}
$
$e=\frac{5}{4}$ and $a=4 \quad \Rightarrow a e=5$ and $\frac{a}{e}=\frac{4}{5 / 4}=\frac{16}{5}$
Now Centre $=(0,0)$
Vertices $=(0, \pm a)=(0, \pm 4)$
Foci $=(0, \pm a)=(0, \pm 5)$
Equation of directrices $y=\pm \frac{16}{5}$

Question 6.
Prove that the length of the latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{2 b^2}{a}$
Solution:
The latus rectum LL' of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through the focus(ae, 0 )
So, $\quad$ Let $\mathrm{L}=(a e, y)$ (say)
$
\begin{aligned}
& \frac{a^2 e^2}{a^2}-\frac{y_1^2}{b^2}=1 \Rightarrow \frac{y_1^2}{b^2}=e^2-1 \\
& \Rightarrow \quad y_1^2=b^2\left(e^2-1\right)=b^2\left(\frac{b^2}{a^2}\right) \\
& \left(\because e^2=1+\frac{b^2}{a^2}\right) \\
& \Rightarrow \quad y_1=\pm \frac{b^2}{a} \\
&
\end{aligned}
$
That is the end point of latus rectum $\mathrm{L}$ and $\mathrm{L}^{\prime}$ are $\left(a e, \frac{b^2}{a}\right)$ and $\left(a e, \frac{-b^2}{a}\right)$.
So the length of the latus rectum $L^{\prime}=\frac{2 b^2}{a}$.

Question 7.
Show that the absolute value of the difference of the focal distances of any point $P$ on the hyperbola is the length of its transverse axis.
Solution:

$\begin{aligned}
& \begin{array}{l}
\text { We know } \frac{\mathrm{SP}}{\mathrm{PM}}=e \\
\Rightarrow \quad \mathrm{SP}=e \mathrm{PM}
\end{array} \\
& \Rightarrow \quad \mathrm{SP}=e\left(x-\frac{a}{e}\right) \\
& =e x-a \\
& \text { Again } \quad \frac{\mathrm{S}^{\prime} \mathrm{P}}{\mathrm{PM}^{\prime}}=e \\
& \Rightarrow \quad \quad \mathrm{S}^{\prime} \mathrm{P}=e \mathrm{PM}^{\prime}=e\left(\frac{a}{e}+x\right) \\
& =a+e x \\
&
\end{aligned}$

$
\begin{aligned}
& \therefore \mathrm{S}^{\prime} \mathrm{P}-\mathrm{SP}=(\mathrm{a}+e \mathrm{ex})-(e \mathrm{ex}-\mathrm{a}) \\
& \mathrm{a}+\mathrm{ex}-\mathrm{ex}+\mathrm{a}=2 \mathrm{a}(\text { transverse axis })
\end{aligned}
$
Question 8.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

(i) $\frac{(x-3)^2}{225}+\frac{(y-4)^2}{289}=1$
(ii) $\frac{(x+1)^2}{100}+\frac{(y-2)^2}{64}=1$
(iii) $\frac{(x+3)^2}{225}-\frac{(y-4)^2}{64}=1$
(iv) $\frac{(y-2)^2}{25}-\frac{(x+1)^2}{16}=1$
(v) $18 x^2+12 y^2-144 x+48 y+120=0$
(vi) $9 x^2-y^2-36 x-6 y+18=0$
Solution:
(i) $\frac{(x-3)^2}{225}+\frac{(y-4)^2}{289}=1$
Now taking $\quad x-3=\mathrm{X}$ and $y-4=\mathrm{Y}$ We get
$
\frac{X^2}{225}+\frac{Y^2}{289}=1
$
It is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, Which is an ellipse
Here major axis is along Y-axis also $a>b$
$\Rightarrow \quad a^2=289$ and $b^2=225(a>b) \Rightarrow a=17$
Now
$
e^2=\frac{a^2-b^2}{a^2}=\frac{289-225}{289}=\frac{64}{289}
$
$\Rightarrow \quad e=\frac{8}{17}$
Now $e=\frac{8}{17}, a=17$
$\Rightarrow a e=8$ and $\frac{a}{e}=\frac{17}{8 / 17}=\frac{289}{18}$

(ii) $\frac{(x+1)^2}{100}+\frac{(y-2)^2}{64}=1$
Taking $\quad x+1=\mathrm{X}$ and $y-2=\mathrm{Y}$ We get
$
\frac{X^2}{100}+\frac{Y^2}{64}=1
$
It is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, Which is an ellipse
Here $a^2=100$ and $b^2=64(a>b)$
$\Rightarrow \quad a=10$ and $b=8$
Now
$e^2=\frac{a^2-b^2}{a^2}=\frac{100-64}{100}=\frac{36}{100}=\frac{9}{25}$
$\Rightarrow \quad e=\frac{3}{5}$
Now $\quad a=10$ and $e=\frac{3}{5}, a e=10 \times \frac{3}{5}=6$ and $\frac{a}{e}=\frac{10}{3 / 5}=\frac{50}{3}$
Here major axis is along $\mathrm{X}$-axis

(iii) $\frac{(x+3)^2}{225}-\frac{(y-4)^2}{64}=1$
Taking $x+3=\mathrm{X}$ and $y-4=\mathrm{Y}$ We get $\frac{\mathrm{X}^2}{225}-\frac{\mathrm{Y}^2}{64}=1$
It is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, Which is a hyperbola
(ie.,) The given conic is a hyperbola
Here $\quad a^2=225, b^2=64$

$
\begin{array}{ll}
\Rightarrow & a=15 \\
\text { Now } & e^2=\frac{a^2+b^2}{a^2}=\frac{225+64}{225}=\frac{289}{225} \\
\Rightarrow & e=\sqrt{\frac{289}{225}}=\frac{17}{15} \\
\text { Now } & a=15, e=\frac{17}{15}, a e=17 \text { and } \frac{a}{e}=\frac{15}{17 / 15}=\frac{225}{17}
\end{array}
$
Here the transverse axis is along $\mathrm{X}$-axis

(iv) $\frac{(y-2)^2}{25}-\frac{(x+1)^2}{16}=1$
Taking $\quad y-2=\mathrm{Y}$ and $x+1=\mathrm{X}$ We get
$
\frac{Y^2}{25}-\frac{X^2}{16}=1
$
It is of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, Which is a hyperbola
Here the transverse axis is along Y-axis
Also $\quad a^2=25, b^2=16$
$
\Rightarrow \quad a=5
$
Now $e^2=\frac{a^2+b^2}{a^2}=\frac{25+16}{25}=\frac{41}{25}$
$
\Rightarrow \quad e=\frac{\sqrt{41}}{5}
$
Now
$
a=5, e=\frac{\sqrt{41}}{5}, a e=\sqrt{41} \text { and } \frac{a}{e}=\frac{5}{\frac{\sqrt{41}}{5}}=\frac{25}{\sqrt{41}}
$

$\begin{aligned}
& \text { (v) } 18 x^2+12 y^2-144 x+48 y+120=0 \\
& \left(18 x^2-144 x\right)+\left(12 y^2+48 y\right)=-120 \\
& 18\left(x^2-8 x\right)+12\left(y^2+4 y\right)=-120 \\
& 18\left(x^2-8 x+16-16\right)+12\left(y^2+4 y+4-4\right)=-120 \\
& 18(x-4)^2-288+12(y+2)^2-48=-120 \\
& 18(x-4)^2+12(y+2)^2=-120+288+48=216
\end{aligned}$

$
\begin{aligned}
\left(\div \text { by 216) } \Rightarrow \frac{18(x-4)^2}{216}+\frac{12(y+2)^2}{216}\right. & =1 \\
\Rightarrow \quad \frac{(x-4)^2}{12}+\frac{(y+2)^2}{18} & =1
\end{aligned}
$
Taking $x-4=\mathrm{X}$ and $y+2=\mathrm{Y}$
We get $\frac{X^2}{12}+\frac{Y^2}{18}=1$
It is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, Which is an ellipse
Here the major axis is along Y-axis
$
\begin{aligned}
& \text { Also } \begin{aligned}
a^2 & =18, b^2=12(a>b) \\
a & =3 \sqrt{2}, b=2 \sqrt{3} \\
e^2 & =\frac{a^2-b^2}{a^2}=\frac{18-12}{18}=\frac{6}{18}=\frac{1}{3} \\
e & =\frac{1}{\sqrt{3}} \\
\text { Now } e & =\frac{1}{\sqrt{3}} \text { and } a=3 \sqrt{2}, \\
a e & =3 \sqrt{2} \frac{1}{\sqrt{3}}=\sqrt{6} \text { and } \frac{a}{e}=\frac{3 \sqrt{2}}{1 / \sqrt{3}}=3 \sqrt{2} \sqrt{3}=3 \sqrt{6}
\end{aligned}
\end{aligned}
$

(vi)
$
\begin{aligned}
9 x^2-y^2-36 x-6 y+18 & =0 \\
\left(9 x^2-36 x\right)-\left(y^2+6 y\right) & =-18 \quad \Rightarrow 9\left(x^2-4 x\right)-\left(y^2+6 y\right)=-18 \\
9\left(x^2-4 x+4-4\right)-\left(y^2+6 y+9-9\right) & =-18 \quad \Rightarrow 9(x-2)^2-36-(y+3)^2+9=-18 \\
9(x-2)^2-(y+3)^2 & =-18+36-9=9 \\
(\div \text { by } 9) \Rightarrow \quad \frac{(x-2)^2}{1}-\frac{(y+3)^2}{9} & =1
\end{aligned}
$
Taking $x-2=\mathrm{X}$ and $y+3=\mathrm{Y}$
We get
$
\frac{X^2}{1}-\frac{Y^2}{9}=1
$

It is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ Which is a hyperbola,
The transverse axis is along X-axis. Here $a^2=1, b^2=9 \Rightarrow a=1$
$
\begin{aligned}
e^2 & =\frac{a^2+b^2}{a^2}=\frac{1+9}{1}=10 \quad \Rightarrow e=\sqrt{10} \\
\text { Now } e & =\sqrt{10} \text { and } a=1, a e=\sqrt{10} \text { and } \frac{a}{e}=\frac{1}{\sqrt{10}}
\end{aligned}
$