Eliminate θ between the given equations:
x = h + a cos θ, y = k + b sin θ
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⇒ x = h + a cos θ
⇒ cos θ = x−ha\dfrac{x - h}{a}ax−h…..(1)
⇒ y = k + b sin θ
⇒ sin θ = y−kb\dfrac{y - k}{b}by−k…..(2)
Square and add equations (1) and (2):
⇒(x−ha)2+(y−kb)2=cos2θ+sin2θ⇒(x−ha)2+(y−kb)2=1⇒(x−h)2a2+(y−k)2b2=1.\Rightarrow \Big(\dfrac{x - h}{a}\Big)^2 + \Big(\dfrac{y - k}{b}\Big)^2 = \cos^2 \theta + \sin^2 \theta \\[1em] \Rightarrow \Big(\dfrac{x - h}{a}\Big)^2 + \Big(\dfrac{y - k}{b}\Big)^2 = 1 \\[1em] \Rightarrow \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 .⇒(ax−h)2+(by−k)2=cos2θ+sin2θ⇒(ax−h)2+(by−k)2=1⇒a2(x−h)2+b2(y−k)2=1.
Hence,the required relation is (x−h)2a2+(y−k)2b2=1\dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1a2(x−h)2+b2(y−k)2=1.
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x = a sec3 θ, y = b tan3 θ
(xa)cosθ+(yb)sinθ=1,(xa)sinθ−(yb)cosθ=−1\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta = 1, \Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta = -1(ax)cosθ+(by)sinθ=1,(ax)sinθ−(by)cosθ=−1
If cosAcosB\dfrac{\cos A}{\cos B}cosBcosA = m and cosAsinB\dfrac{\cos A}{\sin B}sinBcosA = n, prove that : (m2 + n2)cos2B = n2
If x = a sec A cos B, y = b sec A sin B and z = c tan A, prove that x2a2+y2b2−z2c2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1a2x2+b2y2−c2z2=1
