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Question

At the point of intersection of the rectangular hyperbola $$xy=c^2$$ and the parabola $$y^2=4ax$$ tangents to the rectangular hyperbola and the parabola make an angle $$\theta $$ and $$\phi $$ respectively with the axis of $$X$$ , then 


A
θ=tan1(tanϕ)
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B
ϕ=tan1(tanθ)
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C
θ=12tan1(tanϕ)
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D
ϕ=12tan1(tanθ)
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Solution

The correct option is A $$\theta = {\tan ^{ - 1}}\left( { - \tan \phi } \right)$$
Given hyperbola $${ xy=c }^{ 2 }$$ parabola $${ y }^{ 2 }=4ax$$

point of intersection of hyperbola and parabola

$$\\ \Rightarrow { y }^{ 2 }=4a\left( \cfrac { { c }^{ 2 } }{ y }  \right) $$
$$ \Rightarrow { y }^{ 3 }=4a{ c }^{ 2 }$$
$$\Rightarrow y=\sqrt [ 3 ]{ 4a{ c }^{ 2 } } $$
$$\Rightarrow x=\cfrac { { c }^{ 2 } }{ \sqrt [ 3 ]{ 4a{ c }^{ 2 } }  } $$

Point P : $$\left( \cfrac { { c }^{ 2 } }{ \sqrt [ 3 ]{ 4a{ c }^{ 2 } }  }  ,\sqrt [ 3 ]{ 4a{ c }^{ 2 } } \quad  \right) $$

Equation of tangent at $$P$$ on hyperbola, $$\cfrac { x\sqrt [ 3 ]{ 4a{ c }^{ 2 } } +y\left( \cfrac { { c }^{ 2 } }{ \sqrt [ 3 ]{ 4a{ c }^{ 2 } }  }  \right)  }{ 2 } ={ c }^{ 2 }$$

$$\\ \Rightarrow 4a{ c }^{ 2 }x+{ c }^{ 2 }y={ 2c }^{ 2 }\\ \Rightarrow y=-4ax+2$$

Slope = $$\tan { \theta  } $$ =$$ -4a$$----------$$(1)$$

equation of tangent at P on parabola ,
$$y\left( \sqrt [ 3 ]{ 4a{ c }^{ 2 } }  \right) =4a\left[ \left( \cfrac { { c }^{ 2 } }{ \sqrt [ 3 ]{ 4a{ c }^{ 2 } }  }  \right) +x \right] \\ \Rightarrow y=4a\left( \cfrac { { c }^{ 2 } }{ 4a{ c }^{ 2 } } +\cfrac { x }{ \sqrt [ 3 ]{ 4a{ c }^{ 2 } }  }  \right) \\ \Rightarrow y=4ax+\cfrac { { 4ac }^{ 2 } }{ 4a{ c }^{ 2 } } $$

Slope =$$ 4a$$ = $$\tan { \phi  } $$----------------$$(2)$$

From equation $$(1)$$ and $$(2)$$,we get,

$$\tan { \theta  } =-\tan { \phi  } $$

$$\Rightarrow \theta =\tan ^{ -1 }{ \left( -\tan { \phi  }  \right)  } $$

996514_1044552_ans_374cd4fb568b45f09a1435771457b744.png

Maths

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