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Q.Show that the curves y2=4(x+1)y^2 = 4(x + 1) and y2=36(9−x)y^2 = 36(9 - x) intersect orthogonally.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2023Subjective· 7mImportance★★★★★
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Slopes are 2y\frac{2}{y} and −18y\frac{-18}{y}; their product −36y2\frac{-36}{y^2} equals −1-1 at the intersection where y2=36y^2=36.

For y2=4(x+1)y^2=4(x+1): differentiate, 2ydydx=42y\dfrac{dy}{dx}=4, so slope m1=2ym_1=\dfrac{2}{y}.

For y2=36(9−x)y^2=36(9-x): differentiate, 2ydydx=−362y\dfrac{dy}{dx}=-36, so slope m2=−18ym_2=\dfrac{-18}{y}.

Product of slopes: m1m2=2y⋅−18y=−36y2m_1m_2=\dfrac{2}{y}\cdot\dfrac{-18}{y}=\dfrac{-36}{y^2}.

Find the point of intersection: 4(x+1)=36(9−x)⇒4x+4=324−36x⇒40x=320⇒x=84(x+1)=36(9-x)\Rightarrow 4x+4=324-36x\Rightarrow 40x=320\Rightarrow x=8. Then y2=4(8+1)=36y^2=4(8+1)=36.

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