aIn this project, we will calculate the Morse index of FBMS catenoid inside a Spheroid. Consider a catenoid inside and spheroid. It is genus zero and has two boundary components. This surface describe by a conformal map 𝛴=X(M)${\displaystyle 𝛴=X(M)}$ for M${\displaystyle M}$ cylinder [-T,T]×S1${\displaystyle [-T,T]\times S^1}$ with coordinates (t,𝜃)${\displaystyle (t,𝜃)}$. The critical catenoid is isometric to the immersion

X(t,𝜃)

=c(coshtcos𝜃,coshtsin𝜃,t),

$$\begin{array}{rl}X(t,𝜃)& =c(\cosh t\cos 𝜃,\cosh t\sin 𝜃,t),\end{array}$$where 𝜃∈[0,2𝜋)${\displaystyle 𝜃\in [0,2𝜋)}$ and t∈[-T,T]${\displaystyle t\in [-T,T]}$.

Figure 1:X(t,𝜃)${\displaystyle X(t,𝜃)}$ with c=1${\displaystyle c=1}$ and t∈[-1,1]${\displaystyle t\in [-1,1]}$

Now we can calculate tangent vector fields of critical catenoid.

	Xt(t,𝜃)	=c(sinhtcos𝜃,sinhtsin𝜃,1)
	X𝜃(t,𝜃)	=c(-coshtsin𝜃,coshtcos𝜃,0)

$$\begin{array}{rl}{X}_t(t,𝜃)& =c(\sinh t\cos 𝜃,\sinh t\sin 𝜃,1)\\ X_{𝜃}(t,𝜃)& =c(-\cosh t\sin 𝜃,\cosh t\cos 𝜃,0)\end{array}$$

Figure 2:Xt(t,𝜃)${\displaystyle X_t(t,𝜃)}$ with c=1${\displaystyle c=1}$ and t∈[-1,1]${\displaystyle t\in [-1,1]}$

Figure 3:X𝜃(t,𝜃)${\displaystyle X_{𝜃}(t,𝜃)}$ with c=1${\displaystyle c=1}$ and t∈[-1,1]${\displaystyle t\in [-1,1]}$

Figure 4:X(t,𝜃), Xt(t,𝜃), X𝜃(t,𝜃)${\displaystyle X(t,𝜃),X_t(t,𝜃),X_{𝜃}(t,𝜃)}$

Observe that

	Xt×X𝜃	=a e1 e2 e3 csinhtcos𝜃 csinhtsin𝜃 c -ccoshtsin𝜃 ccoshtcos𝜃 0
		=(0-c2coshtcos𝜃)e1-(0+c2coshtsin𝜃)e2+(c2coshtsinhtcos2𝜃+c2coshtsinhtsin2𝜃)e3
		=(-c2coshtcos𝜃,-c2coshtsin𝜃,c2coshtsinht(cos2𝜃+sin2𝜃))
		=(-c2coshtcos𝜃,-c2coshtsin𝜃,c2coshtsinht)(1)

$$\begin{array}{rl}{X}_t\times X_{𝜃}& =\left|\begin{array}{ccc}{e}_1& e_2& e_3\\ c\sinh t\cos 𝜃& c\sinh t\sin 𝜃& c\\ -c\cosh t\sin 𝜃& c\cosh t\cos 𝜃& 0\end{array}\right|\\ & =(0-c^2\cosh t\cos 𝜃)e_1-(0+c^2\cosh t\sin 𝜃)e_2+(c^2\cosh t\sinh t{\cos }^2𝜃+c^2\cosh t\sinh t{\sin }^2𝜃)e_3\\ & =(-c^2\cosh t\cos 𝜃,-c^2\cosh t\sin 𝜃,c^2\cosh t\sinh t({\cos }^2𝜃+{\sin }^2𝜃\left)\right)\\ & =(-c^2\cosh t\cos 𝜃,-c^2\cosh t\sin 𝜃,c^2\cosh t\sinh t)\end{array}$$That implies

	|Xt×X𝜃|	=(-c2coshtcos𝜃)2+(-c2coshtsin𝜃)2+(c2coshtsinht)2
		=c4cosh2tcos2𝜃+c4cosh2tsin2𝜃+c4cosh2tsinh2t
		=c4cosh2tcos2𝜃+c4cosh2tsin2𝜃+c4cosh2tsinh2t
		=c2cosh2t(1+sinh2t)
		=c2cosh4t
		=c2cosh2t(2)

$$\begin{array}{rl}|X_t\times X_{𝜃}|& =\sqrt{(-c^2\cosh t\cos 𝜃{)}^2+(-c^2\cosh t\sin 𝜃{)}^2+(c^2\cosh t\sinh t{)}^2}\\ & =\sqrt{c^4{\cosh }^{2}t{\cos }^2𝜃+c^4{\cosh }^{2}t{\sin }^2𝜃+c^4{\cosh }^{2}t{\sinh }^{2}t}\\ & =\sqrt{c^4{\cosh }^{2}t{\cos }^2𝜃+c^4{\cosh }^{2}t{\sin }^2𝜃+c^4{\cosh }^{2}t{\sinh }^{2}t}\\ & =c^2\sqrt{{\cosh }^{2}t(1+{\sinh }^{2}t)}\\ & =c^2\sqrt{{\cosh }^{4}t}\\ & =c^2{\cosh }^{2}t\end{array}$$ Next, we can calculate normal derivative n1${\displaystyle n_1}$ as

	n1	=Xt×X𝜃 |Xt×X𝜃|
		=1 cosht(-cos𝜃,sin𝜃,sinht)(3)

$$\begin{array}{rl}{n}_1& ={\displaystyle \frac{X_t\times X_{𝜃}}{|X_t\times X_{𝜃}|}}\\ & ={\displaystyle \frac{1}{\cosh t}}(-\cos 𝜃,\sin 𝜃,\sinh t)\end{array}$$ (3) by (1) and (2) We know general equation in spherical coordinate for the Spheroid is

E

=(acos𝜃sinv,asin𝜃sinv,bcosv),

$$\begin{array}{rl}\mathbb{E}& =(a\cos 𝜃\sin v,a\sin 𝜃\sin v,b\cos v),\end{array}$$where v∈[0,𝜋]${\displaystyle v\in [0,𝜋]}$ and a${\displaystyle a}$ and b${\displaystyle b}$ are constants.

Figure 5:E${\displaystyle \mathbb{E}}$ with a=1${\displaystyle a=1}$ and b=2${\displaystyle b=2}$

Now we try to deduce Spheroid equation in cylindrical coordinates. Reason behind that is our catenoid equation is in cylindrical coordinates. Spheroid equation is

x2 a2+y2 a2+z2 b2

=1(4)

$$\begin{array}{rl}{\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{a^2}}+{\displaystyle \frac{z^2}{b^2}}& =1\end{array}$$ In cylindrical coordinate system, we can write

	x	=rcos𝜃
	y	=rsin𝜃
	z	=t

$$\begin{array}{rl}x& =r\cos 𝜃\\ y& =r\sin 𝜃\\ z& =t\end{array}$$From (4), we can write

	r2cos2𝜃 a2+r2sin2𝜃 a2+t2 b2	=1(5)
	⟹r2 a2+t2 b2	=1
	⟹t2 b2	=1-r2 a2
	⟹t	=±b1-r2 a2

$$\begin{array}{rl}{\displaystyle \frac{r^2{\cos }^2𝜃}{a^2}}+{\displaystyle \frac{r^2{\sin }^2𝜃}{a^2}}+{\displaystyle \frac{t^2}{b^2}}& =1\\ ⟹{\displaystyle \frac{r^2}{a^2}}+{\displaystyle \frac{t^2}{b^2}}& =1\\ ⟹{\displaystyle \frac{t^2}{b^2}}& =1-{\displaystyle \frac{r^2}{a^2}}\\ ⟹t& =\pm b\sqrt{1-{\displaystyle \frac{r^2}{a^2}}}\end{array}$$Therefore we can write Spheroid equation E${\displaystyle \mathbb{E}}$ is cylindrical coordinates

E

=(rcos𝜃,rsin𝜃,±b1-r2 a2)

$$\begin{array}{rl}\mathbb{E}& =(r\cos 𝜃,r\sin 𝜃,\pm b\sqrt{1-{\displaystyle \frac{r^2}{a^2}}})\end{array}$$

Figure 6:E${\displaystyle \mathbb{E}}$

So in yz${\displaystyle yz}$-plane we have the projection of E${\displaystyle \mathbb{E}}$ as Ep${\displaystyle {\mathbb{E}}_p}$, (set 𝜃=𝜋/2${\displaystyle 𝜃=𝜋/2}$)

Ep

=(r,±b1-r2 a2)

$$\begin{array}{rl}{\mathbb{E}}_p& =(r,\pm b\sqrt{1-{\displaystyle \frac{r^2}{a^2}}})\end{array}$$

Figure 7:Ep${\displaystyle {\mathbb{E}}_p}$ with a=1${\displaystyle a=1}$, b=2${\displaystyle b=2}$

Or I can write this way. From (5) I have

	r2 a2+t2 b2	=1
	r	=±a1-t2 b2
	r	=a1-t2 b2(6)

$$\begin{array}{rl}{\displaystyle \frac{r^2}{a^2}}+{\displaystyle \frac{t^2}{b^2}}& =1\\ r& =\pm a\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}\\ r& =a\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}\end{array}$$(6) since r>0${\displaystyle r>0}$. Therefore we can write Spheroid equation E${\displaystyle \mathbb{E}}$ is cylindrical coordinates

E

=(a1-t2 b2cos𝜃,a1-t2 b2sin𝜃,t)

$$\begin{array}{rl}\mathbb{E}& =(a\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}\cos 𝜃,a\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}\sin 𝜃,t)\end{array}$$

Figure 8:E${\displaystyle \mathbb{E}}$

So in yz${\displaystyle yz}$-plane we have the projection of E${\displaystyle \mathbb{E}}$ as Ep${\displaystyle {\mathbb{E}}_p}$, (set 𝜃=𝜋/2${\displaystyle 𝜃=𝜋/2}$ and 𝜃=-𝜋/2${\displaystyle 𝜃=-𝜋/2}$)

	Ep	=(±a1-t2 b2,t)(7)
	⟹Ept	=(∓at b21 1-t2 b2,1)(8)

$$\begin{array}{rl}{\mathbb{E}}_p& =(\pm a\sqrt{1-{\displaystyle \frac{t^2}{b^2}}},t)\\ ⟹{\mathbb{E}}_{p_t}& =(\mp {\displaystyle \frac{at}{b^2}}{\displaystyle \frac{1}{\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}}},1)\end{array}$$

Figure 9:Ep${\displaystyle {\mathbb{E}}_p}$

Also, projection of Catenoid to yz${\displaystyle yz}$-plane is

Xp

=c(cosht,t)

$$\begin{array}{rl}{X}_p& =c(\cosh t,t)\end{array}$$

Figure 10:Xp${\displaystyle X_p}$ at c=1${\displaystyle c=1}$

Here we can see that, to intersect both catenoid and spheroid, |c|<|a|${\displaystyle |c|<|a|}$.

Figure 11:Ep${\displaystyle {\mathbb{E}}_p}$ and Xp${\displaystyle X_p}$ at c=0.8${\displaystyle c=0.8}$.

We can find tangent vector fields of Xp${\displaystyle X_p}$ as

Xpt

=c(sinht,1)

$$\begin{array}{rl}{X}_{p_t}& =c(\sinh t,1)\end{array}$$

Figure 12:Vector plot of Xpt${\displaystyle X_{p_t}}$ at c=0.8${\displaystyle c=0.8}$

Now I need to find normal vector for Ep${\displaystyle {\mathbb{E}}_p}$. To do that, first I'm trying to find the curvature. Let us take the curve equation of the Ep${\displaystyle {\mathbb{E}}_p}$ is 𝛾Ep${\displaystyle {𝛾}_{{\mathbb{E}}_p}}$. Therefore

𝛾Ep

=(±a1-t2 b2,t)

$$\begin{array}{rl}{𝛾}_{{\mathbb{E}}_p}& =(\pm a\sqrt{1-{\displaystyle \frac{t^2}{b^2}}},t)\end{array}$$Also, we can write

	𝛾'Ep	=(∓at b21 1-t2 b2,1)
		=(∓at b2b b2-t2,1)
		=(∓at b1 b2-t2,1)
	⟹|𝛾'Ep|	=a2t2 b2(b2-t2)+1
		=a2t2+b4-b2t2 b2(b2-t2)
		=t2(a2-b2)+b4 b2(b2-t2)

$$\begin{array}{rl}𝛾{\text{'}}_{{\mathbb{E}}_p}& =(\mp {\displaystyle \frac{at}{b^2}}{\displaystyle \frac{1}{\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}}},1)\\ & =(\mp {\displaystyle \frac{at}{b^{\cancel{2}}}}{\displaystyle \frac{\cancel{b}}{\sqrt{b^2-t^2}}},1)\\ & =(\mp {\displaystyle \frac{at}{b}}{\displaystyle \frac{1}{\sqrt{b^2-t^2}}},1)\\ ⟹|𝛾{\text{'}}_{{\mathbb{E}}_p}|& =\sqrt{{\displaystyle \frac{a^2{t}^2}{b^2(b^2-t^2)}}+1}\\ & =\sqrt{{\displaystyle \frac{a^2{t}^2+b^4-b^2{t}^2}{b^2(b^2-t^2)}}}\\ & =\sqrt{{\displaystyle \frac{t^2(a^2-b^2)+b^4}{b^2(b^2-t^2)}}}\end{array}$$Then arc length function is

	s(t)	=t∫0t2(a2-b2)+b4 b2(b2-t2) dt
		=1 bt∫0t2(a2-b2)+b4 b2-t2 dt

$$\begin{array}{rl}s(t)& =\underset{0}{\overset{t}{\int }}\sqrt{{\displaystyle \frac{t^2(a^2-b^2)+b^4}{b^2(b^2-t^2)}}}dt\\ & ={\displaystyle \frac{1}{b}}\underset{0}{\overset{t}{\int }}\sqrt{{\displaystyle \frac{t^2(a^2-b^2)+b^4}{b^2-t^2}}}dt\end{array}$$Let

	t	=bsin𝜃
	dt	=bcos𝜃 d𝜃

$$\begin{array}{rl}t& =b\sin 𝜃\\ dt& =b\cos 𝜃d𝜃\end{array}$$and t→0, 𝜃→0${\displaystyle t\to 0,𝜃\to 0}$ and t→t, 𝜃→arcsin(t

b)${\displaystyle t\to t,𝜃\to \arcsin \left({\displaystyle \frac{t}{b}}\right)}$. So

	s(t)	=1 barcsin(t b)∫0b2sin2𝜃(a2-b2)+b42 b21-b2sin2𝜃bcos𝜃 d𝜃
		=arcsin(t b)∫0sin2𝜃(a2-b2)+b2 d𝜃
		=barcsin(t b)∫01-(1-a2 b2)sin2𝜃 d𝜃
		=b E(arcsin(t b)|(1-a2 b2))

$$\begin{array}{rl}s(t)& ={\displaystyle \frac{1}{\cancel{b}}}\underset{0}{\overset{\arcsin \left({\displaystyle \frac{t}{b}}\right)}{\int }}\sqrt{{\displaystyle \frac{\cancel{b^2}{\sin }^2𝜃(a^2-b^2)+b^{\cancel{4}2}}{\cancel{\cancel{b^2}1-\cancel{b^2}{\sin }^2𝜃}}}}\cancel{b}\cancel{\cos 𝜃}d𝜃\\ & =\underset{0}{\overset{\arcsin \left({\displaystyle \frac{t}{b}}\right)}{\int }}\sqrt{{\sin }^2𝜃(a^2-b^2)+b^2}d𝜃\\ & =b\underset{0}{\overset{\arcsin \left({\displaystyle \frac{t}{b}}\right)}{\int }}\sqrt{1-(1-{\displaystyle \frac{a^2}{b^2}}){\sin }^2𝜃}d𝜃\\ & =bE\left(\arcsin \right({\displaystyle \frac{t}{b}})|(1-{\displaystyle \frac{a^2}{b^2}}))\end{array}$$, where E(arcsin(t

b)|(1-a2

b2))${\displaystyle E\left(\arcsin \right({\displaystyle \frac{t}{b}})|(1-{\displaystyle \frac{a^2}{b^2}}))}$ is elliptic integral of the second kind. Consider,

	Ep⋅Ept	=(±a1-t2 b2,t)⋅(∓at b21 1-t2 b2,1)
		=-a2t b2+t

$$\begin{array}{rl}{\mathbb{E}}_p\cdot {\mathbb{E}}_{p_t}& =(\pm a\sqrt{1-{\displaystyle \frac{t^2}{b^2}}},t)\cdot (\mp {\displaystyle \frac{at}{b^2}}{\displaystyle \frac{1}{\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}}},1)\\ & =-{\displaystyle \frac{a^{2}t}{b^2}}+t\end{array}$$This proves position vector is not normal to the curve. Next I'm trying to find unit tangent vector to Ep${\displaystyle {\mathbb{E}}_p}$. We know that, from (8),

	Ept	=(∓at b21 1-t2 b2,1)
		=(∓at bb2-t2,1)

$$\begin{array}{rl}{\mathbb{E}}_{p_t}& =(\mp {\displaystyle \frac{at}{b^2}}{\displaystyle \frac{1}{\sqrt{1-{\displaystyle \frac{t^2}{b^2}}}}},1)\\ & =(\mp {\displaystyle \frac{at}{b\sqrt{b^2-t^2}}},1)\end{array}$$Therefore

	T	=Ept |Ept|
		=(∓at bb2-t2,1) a2t2 b2(b2-t2)+1
		=(∓at bb2-t2,1) a2t2-b2t2+b4 b2(b2-t2)
		=(∓at bb2-t2,1) (a2-b2)t2+b4 b2(b2-t2)
		=(∓at bb2-t2,1) (a2-b2)t2+b4 bb2-t2
		=(∓at (a2-b2)t2+b4,bb2-t2 (a2-b2)t2+b4)
	⟹dT dt	=(∓a(a2-b2)t2+b4-at2t(a2-b2) 2(a2-b2)t2+b4 (a2-b2)t2+b4,b(-2t 2b2-t2)(a2-b2)t2+b4-bb2-t2(2(a2-b2)t 2(a2-b2)t2+b4) (a2-b2)t2+b4)
		=(∓a[(a2-b2)t2+b4-t2(a2-b2)] ((a2-b2)t2+b4)3/2,-bt((a2-b2)t2+b4)-b(b2-t2)(a2-b2)t ((a2-b2)t2+b4)3/2b2-t2)
		=(∓ab4 ((a2-b2)t2+b4)3/2,-bt((a2-b2)t2+b4+(b2-t2)(a2-b2)) ((a2-b2)t2+b4)3/2b2-t2)
		=(∓ab4 ((a2-b2)t2+b4)3/2,-bt((a2-b2)t2+b4+(b2-t2)(a2-b2)) ((a2-b2)t2+b4)3/2b2-t2)
		=(∓ab4 ((a2-b2)t2+b4)3/2,-bt((a2-b2)t2+b4+(a2-b2)b2-(a2-b2)t2) ((a2-b2)t2+b4)3/2b2-t2)
		=(∓ab4 ((a2-b2)t2+b4)3/2,-bt(b4+a2b2-b4) ((a2-b2)t2+b4)3/2b2-t2)
		=(∓ab4 ((a2-b2)t2+b4)3/2,-a2b3t ((a2-b2)t2+b4)3/2b2-t2)(9)
	⟹adT dta	=a2b8 ((a2-b2)t2+b4)3+a4b6t2 ((a2-b2)t2+b4)3(b2-t2)
		=a2b8(b2-t2) ((a2-b2)t2+b4)3(b2-t2)+a4b6t2 ((a2-b2)t2+b4)3(b2-t2)
		=a2b8(b2-t2)+a4b6t2 ((a2-b2)t2+b4)3(b2-t2)
		=a2b10-a2b8t2+a4b6t2 ((a2-b2)t2+b4)3(b2-t2)
		=a2b10-t2a2b6(b2-a2) ((a2-b2)t2+b4)3(b2-t2)
		=ab3b4+t2(-b2+a2) ((a2-b2)t2+b4)32(b2-t2)
		=ab3 ((a2-b2)t2+b4)b2-t2(10)

$$\begin{array}{rl}T& ={\displaystyle \frac{{\mathbb{E}}_{p_t}}{|{\mathbb{E}}_{p_t}|}}\\ & ={\displaystyle \frac{(\mp {\displaystyle \frac{at}{b\sqrt{b^2-t^2}}},1)}{\sqrt{{\displaystyle \frac{a^2{t}^2}{b^2(b^2-t^2)}}+1}}}\\ & ={\displaystyle \frac{(\mp {\displaystyle \frac{at}{b\sqrt{b^2-t^2}}},1)}{\sqrt{{\displaystyle \frac{a^2{t}^2-b^2{t}^2+b^4}{b^2(b^2-t^2)}}}}}\\ & ={\displaystyle \frac{(\mp {\displaystyle \frac{at}{b\sqrt{b^2-t^2}}},1)}{\sqrt{{\displaystyle \frac{(a^2-b^2)t^2+b^4}{b^2(b^2-t^2)}}}}}\\ & ={\displaystyle \frac{(\mp {\displaystyle \frac{at}{b\sqrt{b^2-t^2}}},1)}{{\displaystyle \frac{\sqrt{(a^2-b^2)t^2+b^4}}{b\sqrt{b^2-t^2}}}}}\\ & =(\mp {\displaystyle \frac{at}{\sqrt{(a^2-b^2)t^2+b^4}}},{\displaystyle \frac{b\sqrt{b^2-t^2}}{\sqrt{(a^2-b^2)t^2+b^4}}})\\ ⟹{\displaystyle \frac{dT}{dt}}& =(\mp {\displaystyle \frac{a\sqrt{(a^2-b^2)t^2+b^4}-at{\displaystyle \frac{\cancel{2}t(a^2-b^2)}{\cancel{2}\sqrt{(a^2-b^2)t^2+b^4}}}}{(a^2-b^2)t^2+b^4}},{\displaystyle \frac{b\left({\displaystyle \frac{-\cancel{2}t}{\cancel{2}\sqrt{b^2-t^2}}}\right)\sqrt{(a^2-b^2)t^2+b^4}-b\sqrt{b^2-t^2}\left({\displaystyle \frac{\cancel{2}(a^2-b^2)t}{\cancel{2}\sqrt{(a^2-b^2)t^2+b^4}}}\right)}{(a^2-b^2)t^2+b^4}})\\ & =(\mp {\displaystyle \frac{a[\cancel{(a^2-b^2)t^2}+b^4-\cancel{t^2(a^2-b^2)}]}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}}},{\displaystyle \frac{-bt\left(\right(a^2-b^2)t^2+b^4)-b(b^2-t^2)(a^2-b^2)t}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}\sqrt{b^2-t^2}}})\\ & =(\mp {\displaystyle \frac{a{b}^4}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}}},{\displaystyle \frac{-bt\left(\right(a^2-b^2)t^2+b^4+(b^2-t^2\left)\right(a^2-b^2\left)\right)}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}\sqrt{b^2-t^2}}})\\ & =(\mp {\displaystyle \frac{a{b}^4}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}}},{\displaystyle \frac{-bt\left(\right(a^2-b^2)t^2+b^4+(b^2-t^2\left)\right(a^2-b^2\left)\right)}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}\sqrt{b^2-t^2}}})\\ & =(\mp {\displaystyle \frac{a{b}^4}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}}},{\displaystyle \frac{-bt(\cancel{(a^2-b^2)t^2}+b^4+(a^2-b^2)b^2-\cancel{(a^2-b^2)t^2})}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}\sqrt{b^2-t^2}}})\\ & =(\mp {\displaystyle \frac{a{b}^4}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}}},{\displaystyle \frac{-bt(\cancel{b^4}+a^2{b}^2-\cancel{b^4})}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}\sqrt{b^2-t^2}}})\\ & =(\mp {\displaystyle \frac{a{b}^4}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}}},{\displaystyle \frac{-a^2{b}^{3}t}{\left(\right(a^2-b^2)t^2+b^4{)}^{3/2}\sqrt{b^2-t^2}}})\\ ⟹\left|{\displaystyle \frac{dT}{dt}}\right|& =\sqrt{{\displaystyle \frac{a^2{b}^8}{\left(\right(a^2-b^2)t^2+b^4{)}^3}}+{\displaystyle \frac{a^4{b}^6{t}^2}{\left(\right(a^2-b^2)t^2+b^4{)}^3(b^2-t^2)}}}\\ & =\sqrt{{\displaystyle \frac{a^2{b}^8(b^2-t^2)}{\left(\right(a^2-b^2)t^2+b^4{)}^3(b^2-t^2)}}+{\displaystyle \frac{a^4{b}^6{t}^2}{\left(\right(a^2-b^2)t^2+b^4{)}^3(b^2-t^2)}}}\\ & =\sqrt{{\displaystyle \frac{a^2{b}^8(b^2-t^2)+a^4{b}^6{t}^2}{\left(\right(a^2-b^2)t^2+b^4{)}^3(b^2-t^2)}}}\\ & =\sqrt{{\displaystyle \frac{a^2{b}^{10}-a^2{b}^8{t}^2+a^4{b}^6{t}^2}{\left(\right(a^2-b^2)t^2+b^4{)}^3(b^2-t^2)}}}\\ & =\sqrt{{\displaystyle \frac{a^2{b}^{10}-t^2{a}^2{b}^6(b^2-a^2)}{\left(\right(a^2-b^2)t^2+b^4{)}^3(b^2-t^2)}}}\\ & =a{b}^3\sqrt{{\displaystyle \frac{\cancel{b^4+t^2(-b^2+a^2)}}{\left(\right(a^2-b^2)t^2+b^4{)}^{\cancel{3}2}(b^2-t^2)}}}\\ & ={\displaystyle \frac{a{b}^3}{\left(\right(a^2-b^2)t^2+b^4)\sqrt{b^2-t^2}}}\end{array}$$ Then unit normal vector can be written as N=dT

dt

|dT

dt|${\displaystyle N={\displaystyle \frac{{\displaystyle \frac{dT}{dt}}}{|{\displaystyle \frac{dT}{dt}}|}}}$. So by (9), (10) and definition of N${\displaystyle N}$,

	N	=(∓ab4 ((a2-b2)t2+b4)31/2,-a2b3t ((a2-b2)t2+b4)31/2b2-t2) ab3 ((a2-b2)t2+b4)b2-t2
		=(∓bb2-t2 ((a2-b2)t2+b4)1/2,-at ((a2-b2)t2+b4)1/2)
		=(∓bb2-t2 (a2-b2)t2+b4,-at (a2-b2)t2+b4)

$$\begin{array}{rl}N& ={\displaystyle \frac{(\mp {\displaystyle \frac{\cancel{a}b\cancel{{}^4}}{\left(\right(a^2-b^2)t^2+b^4{)}^{\cancel{3}1/2}}},{\displaystyle \frac{-a\cancel{{}^2}\cancel{b^3}t}{\left(\right(a^2-b^2)t^2+b^4{)}^{\cancel{3}1/2}\sqrt{b^2-t^2}}})}{{\displaystyle \frac{\cancel{a}\cancel{b^3}}{\cancel{\left(\right(a^2-b^2)t^2+b^4)}\sqrt{b^2-t^2}}}}}\\ & =(\mp {\displaystyle \frac{b\sqrt{b^2-t^2}}{\left(\right(a^2-b^2)t^2+b^4{)}^{1/2}}},-{\displaystyle \frac{at}{\left(\right(a^2-b^2)t^2+b^4{)}^{1/2}}})\\ & =(\mp {\displaystyle \frac{b\sqrt{b^2-t^2}}{\sqrt{(a^2-b^2)t^2+b^4}}},-{\displaystyle \frac{at}{\sqrt{(a^2-b^2)t^2+b^4}}})\end{array}$$ ${\displaystyle }$