Question

$x = 2a \sec( \alpha ) y = 2b \tan( \alpha ) then \: {x}^{2} \div {a}^{2} - {y}^{2} \div {b}^{2} =$
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Answer 1

Question :-

$\sf \: x = 2a \sec( \alpha ) , \: y = 2b \tan( \alpha ), then \: find \\ \sf\frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } =$

Answer :-

$\to \boxed{ \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 4} \:$

Solution :-

We have ,

$\mapsto \sf \: x = 2a \sec (\alpha) \: ....... eq.(1)\\ \\ \mapsto \sf y = 2b \tan( \alpha ) \: .......eq.(2)$

Method (1)

Firstly squaring on both sides in both eq.

$\to \: \sf {x}^{2} = { \{ 2a \sec (\alpha) \: \}}^{2} \\ \\ \to \sf {x}^{2} =4 {a}^{2} \: { \sec}^{2}( \alpha) \: \: ....eq.(3) \\ \bf and \\ \\ \to \sf {y}^{2} = { \{ 2b \tan (\alpha) \} }^{2} \: \\ \\ \to \sf \: {y}^{2} = 4 {b}^{2} { \tan}^{2}( \alpha) \: \: .....eq.(4)$

Now,to eq.(3) divided by a² ,and to eq.(4) divided by b² after that subtract them

$\to \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = \frac{4 {a}^{2} { \sec}^{2}( \alpha) }{ {a}^{2} } - \frac{4 {b}^{2} { \tan}^{2}( \alpha) }{ {b}^{2} } \\ \\ \to \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 4 \{ { \sec}^{2} ( \alpha) - { \tan}^{2} ( \alpha) \} \\ \\ \because \: { \sec}^{2} \theta - { \tan}^{2} \theta = 1 \\ \\ \to \boxed{ \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 4}$

Method (2)

We have ,

$\mapsto \sf \: x = 2a \sec (\alpha) \: \\ \\ \mapsto \sf y = 2b \tan( \alpha ) \: \\ \: \\ \sf \: hence \: \\ \\ \to \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } \\ \\ \to \sf \frac{ \{ {2a \sec (\alpha) \}}^{2} }{ {a}^{2} } - \frac{ { \{2b \tan( \alpha ) \}}^{2} }{ {b}^{2} } \\ \\ \to \sf \frac{4 {a}^{2} { \sec}^{2} ( \alpha) }{ {a}^{2} } - \frac{4 {b}^{2} { \tan}^{2}( \alpha) }{ {b}^{2} } \\ \\ \to \: 4 \big \{ { \sec}^{2} \alpha - { \tan}^{2} \alpha \big \} \\ \\ \to \: 4$