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

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

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