Question

Prove that b^2.x^2-a^2.y^2=a^2.b^2
If (1). x=a sec theta , y= b tan theta

Answer 1

Solution :-

Given :-

• x = a . sec θ

• y = b . tan θ

x = a . sec θ

⇒ x / a = sec θ

Squaring on both sides

⇒ ( x / a )² = sec² θ

⇒ x² / a² = sec² θ -- eq( 1 )

y = b . tan θ

⇒ y / b = tan θ

Squaring on both sides

⇒ ( y / b )² = tan² θ

⇒ y² / b² = tan² θ -- eq ( 2 )

$\tt Subtracting \ eq(2) \ from \ eq(1) \\ \\ \implies \dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = \sec^{2} \theta - \tan^{2} \theta \\ \\ \implies\dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \qquad \{ \because \sec^{2} \theta - \tan^{2} \theta = 1 \} \\ \\ \tt Multiplying \ thoughout \ by \ {a}^{2} {b}^{2} \\ \\ \implies \dfrac{ {x}^{2} }{ {a}^{2} } ( {a}^{2} {b}^{2} ) - \dfrac{ {y}^{2} }{ {b}^{2} } ( {a}^{2} {b}^{2} ) = 1( {a}^{2} {b}^{2}) \\ \\ \implies {b}^{2} {x}^{2} - {a}^{2} {y}^{2} = {a}^{2} {b}^{2} \\ \\ \bf Hence \ proved.$