Eliminate θ from the equation xsinθ−ycosθ=x2+y2 and cos2θa2+sin2θb2=1x2+y2

xsinθ−ycosθ=x2+y2→(1)andcos2θa2+sin2θb2=1x2+y2→(2)Squaring on both sides of (i)x2sin2θ+y2cos2θ−2xysinθ.cosθ=x2+y2x2(1−cos2θ)+y2(1−sin2θ)−2xysinθ.cosθ=x2+y2x2−x2cos2θ+y2−y2sin2θ−2xysinθ.cosθ=x2+y2x2+y2−(xcosθ+ysinθ)2=x2+y2(xcosθ+ysinθ)2=0xcosθ=−ysinθtanθ=−xy tanθ is negative in Q2 & Q4Case(1): θ∈Q2∴cosθ=−yx2+y2 and sinθ=xx2+y2Substituting these values of cosθ and sinθ in (ii)1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2⇒x2a2+y2a2=1Case(2): θ∈Q4∴cosθ=−yx2+y2 and sinθ=−xx2+y2Substituting these values of cosθ and sinθ in (ii)1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2⇒x2b2+y2a2=1

Eliminate θ from the equation xsinθ−ycosθ=x2+y2 and cos2θa2+sin2θb2=1x2+y2

xsinθ−ycosθ=x2+y2→(1)andcos2θa2+sin2θb2=1x2+y2→(2)Squaring on both sides of (i)x2sin2θ+y2cos2θ−2xysinθ.cosθ=x2+y2x2(1−cos2θ)+y2(1−sin2θ)−2xysinθ.cosθ=x2+y2x2−x2cos2θ+y2−y2sin2θ−2xysinθ.cosθ=x2+y2x2+y2−(xcosθ+ysinθ)2=x2+y2(xcosθ+ysinθ)2=0xcosθ=−ysinθtanθ=−xy tanθ is negative in Q2 & Q4Case(1): θ∈Q2∴cosθ=−yx2+y2 and sinθ=xx2+y2Substituting these values of cosθ and sinθ in (ii)1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2⇒x2a2+y2a2=1Case(2): θ∈Q4∴cosθ=−yx2+y2 and sinθ=−xx2+y2Substituting these values of cosθ and sinθ in (ii)1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2⇒x2b2+y2a2=1

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Eliminate $θ$ from the equation $x \sin θ - y \cos θ = \sqrt{x^{2} + y^{2}}$ and $\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}} = \frac{1}{x^{2} + y^{2}}$

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$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} = 1$

$\frac{x}{a} + \frac{y}{b} = 1$

$\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$

answer is D.

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Detailed Solution

$\begin{array}{l} x \sin θ - y \cos θ = \sqrt{x^{2} + y^{2}} \to (1) \\ a n d \\ \frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}} = \frac{1}{x^{2} + y^{2}} \to (2) \\ Squaring on both sides of (i) \\ x^{2} \sin^{2} θ + y^{2} \cos^{2} θ - 2 x y \sin θ . \cos θ = x^{2} + y^{2} \\ x^{2} (1 - \cos^{2} θ) + y^{2} (1 - \sin^{2} θ) - 2 x y \sin θ . \cos θ = x^{2} + y^{2} \\ x^{2} - x^{2} \cos^{2} θ + y^{2} - y^{2} \sin^{2} θ - 2 x y \sin θ . \cos θ = x^{2} + y^{2} \\ x^{2} + y^{2} - {(x \cos θ + y \sin θ)}^{2} = x^{2} + y^{2} \\ {(x \cos θ + y \sin θ)}^{2} = 0 \\ x \cos θ = - y \sin θ \\ \tan θ = \frac{- x}{y} \end{array}$

$\begin{array}{l} \tan θ is negative in Q_{2} & Q_{4} \\ C a s e (1) : θ \in Q_{2} \\ ∴ \cos θ = \frac{- y}{\sqrt{x^{2} + y^{2}}} a n d \sin θ = \frac{x}{\sqrt{x^{2} + y^{2}}} \\ Substituting these values of \cos θ a n d \sin θ in (ii) \\ \frac{1}{a^{2}} \times \frac{y^{2}}{x^{2} + y^{2}} + \frac{1}{b^{2}} \times \frac{x^{2}}{x^{2} + y^{2}} = \frac{1}{x^{2} + y^{2}} \\ \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} = 1 \\ C a s e (2) : θ \in Q_{4} \\ ∴ \cos θ = \frac{- y}{\sqrt{x^{2} + y^{2}}} a n d \sin θ = \frac{- x}{\sqrt{x^{2} + y^{2}}} \\ Substituting these values of \cos θ a n d \sin θ in (ii) \\ \frac{1}{a^{2}} \times \frac{y^{2}}{x^{2} + y^{2}} + \frac{1}{b^{2}} \times \frac{x^{2}}{x^{2} + y^{2}} = \frac{1}{x^{2} + y^{2}} \\ \Rightarrow \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \end{array}$