Question

If :
$\boxed{ \begin{array}{cc}\sf \bull \: x = asec \theta \: cos \phi \\ \\ \sf\bull \: y = bsec \theta sin \phi \: \: \\ \\ \sf \bull \: z = ctan \theta \: \: \: \: \: \: \: \: \: \: \end{array}}$
⠀
Then find the value of :
$\boxed{ \sf \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } }$
Please answer with explanation. :) ​

Answer 1

Aиѕωєr —

ㅤ

• $\tt{ \red{\leadsto} \boxed{ \: \: \: \left(1+\dfrac{z^2}{c^2}\right) \: \: }}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

Giνєи —

$\qquad{ \large {\maltese \: } \small {\boxed{ \begin{array}{cc} \sf{\bull \: x = a \: sec \theta cos \theta } \\ \sf{ \bull \: y = b \: sec \theta sin \theta \: } \\ \sf{ \bull \: z = c \: tan \theta \: \: \: \: \: \: \: \: } \\ \end{array}}}}$

ㅤ

Tσ Fiиd —

• $\sf{\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\right)}$

ㅤ

Sσℓυтiσи –

✪ We have :

$\quad\tt{\longrightarrow x = a\: sec\theta cos\theta }$

$\quad\tt{\longrightarrow \dfrac{x}{a}=sec\theta cos \theta }$

ㅤ

✪ On squaring both sides we get :

$\quad\tt{\longrightarrow \dfrac{x^2}{a^2}=sec^2 \theta cos^2 \theta \qquad \:\:\:\:\:\:\:\:\:\:\: \cdots\cdots ( 1 ) }$

ㅤ

✪ Now :

$\quad\tt{\longrightarrow y = b \: sec \theta sin \theta }$

$\quad\tt{\longrightarrow \dfrac{y}{b}= sec\theta sin\theta }$

ㅤ

✪ On squaring both sides :

$\quad\tt{\longrightarrow \dfrac{y^2}{b^2}=sec^2 \theta sin^2 \theta \qquad \:\:\:\:\:\:\:\:\:\:\: \cdots \cdots ( 2 )}$

ㅤ

✪ Now :

$\quad\tt{\longrightarrow z = c \: tan \theta }$

$\quad\tt{\longrightarrow \dfrac{z}{c}=tan\theta }$

ㅤ

✪ On squaring both sides :

$\quad\tt{\longrightarrow \dfrac{z^2}{c^2} = tan^2 \theta \qquad \:\:\:\:\:\:\:\:\:\:\:\cdots \cdots ( 3 ) }$

ㅤ

ㅤ✪ On adding equation ( 1 ) and ( 2 ) :

$\quad\tt{\longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2 }=sec^2 \theta cos^2 \theta + sec^2 \theta sin^2\theta }$

$\quad\tt{\longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2 }=sec^2 \theta ( cos^2 \theta + sin^2 \theta )\qquad \qquad\:\:\:\:\:\:\:\:\:\:\: \bigg\lgroup cos ^2 \theta + sin^2 \theta = 1 \bigg\rgroup}$

$\quad\tt{\longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2 }= sec ^2 \theta \times 1 }$

$\quad\tt{\longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2 }= sec^2 \theta }$

$\quad\tt{\longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2 }= tan^2 \theta +1 \qquad\qquad \:\:\:\:\:\:\:\:\:\:\:\bigg\lgroup sec^2 \theta = tan^2 \theta +1 \bigg\rgroup }$

ㅤ

✪ Using equation ( 3 ) :

$\quad\tt{\longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2 }=\dfrac{z^2}{c^2}+1 }$

$\quad{\pmb{\tt{\longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2 }=1+\dfrac{z^2}{c^2}}} }$

ㅤ

❝ Hєиcє Sσℓved ❞

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Answer 2

Step-by-step explanation:

$\[ \begin{array}{l} \color{green} \tt \: Given \\ \\ \color{red}\tt \[ x=a \sec \theta \cos \phi, y=b \sec \theta \sin \phi \] \\ \\ \color{orange}\text{To find the value of \: \: \( \tt\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}} \)} \\ \\ \tt \:\color{navy} Simplifying \\ \\\color{maroon} \tt \: x=a \sec \theta \cos \phi, y=b \sec \theta \sin \phi \\ \\ \color{purple}\tt\text { so }, \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=\dfrac{a^{2} \sec ^{2} \theta \cos ^{2} \phi}{a^{2}}+\dfrac{b^{2} \sec ^{2} \theta \sin ^{2} \phi}{b^{2}} \\ \\ \color{blue}\tt =\sec ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right) \\ \\ \color{darkorange}\tt =\sec ^{2} \theta \end{array} \]$