if we eliminate " θ" from x=asec θ, y=btan θ then we get
a)b[tex]b^{2} x^{2} -a^{2} y^{2} =a^{2} b^{2}[/tex]
b)[tex]b^{2} x^{2} +a^{2} y^{2} =a^{2} b^{2}[/tex]
c)[tex]x^{2} +y^{2} =a^{2} +b^{2}[/tex]
d)[tex]x^{2} -y^{2} =a^{2} -b^{2}[/tex]
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Answer:
[tex]\boxed{\bf\:(a) \: \: \: {b}^{2} {x}^{2} - {a}^{2} {y}^{2} = {a}^{2} {b}^{2} \: } \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf\: x = a \: sec \theta \: \implies\sf\:sec\theta = \dfrac{x}{a} - - - (1) \\ [/tex]
Further given that,
[tex]\sf\: y = b \: tan \theta \: \implies\sf\:tan\theta = \dfrac{y}{b} - - - (2) \\ [/tex]
We know,
[tex]\sf\: {sec}^{2}\theta - {tan}^{2}\theta = 1 \\ [/tex]
On substituting the values from equation (1) and (2), we get
[tex]\sf\: {\bigg(\dfrac{x}{a} \bigg) }^{2} - {\bigg(\dfrac{y}{b} \bigg) }^{2} = 1 \\ [/tex]
[tex]\sf\:\dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \\ [/tex]
[tex]\sf\:\dfrac{ {b}^{2} {x}^{2} - {a}^{2} {y}^{2} }{ {a}^{2} {b}^{2} } = 1 \\ [/tex]
[tex]\implies\sf\: {b}^{2} {x}^{2} - {a}^{2} {y}^{2} = {a}^{2} {b}^{2} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: {b}^{2} {x}^{2} - {a}^{2} {y}^{2} = {a}^{2} {b}^{2} \: } \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]