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Question

Two equal parabolas have the same vertex and their axes are at right angles ; prove that the common tangent touches each at the end of a latus rectum.


Solution

Let the two parabolas be $${ y }^{ 2 }=4ax......(i)$$ and $${ x }^{ 2 }=4ay......(ii)$$

Equation of tangent to $$(i)$$ is

$$y=mx+\dfrac { a }{ m } $$

It also touches $$(ii)$$ so substituting $$y$$ in $$(ii)$$

$${ x }^{ 2 }=4a\left( mx+\dfrac { a }{ m }  \right) \\ m{ x }^{ 2 }=4a{ m }^{ 2 }x+4{ a }^{ 2 }\\ m{ x }^{ 2 }-4a{ m }^{ 2 }x-4{ a }^{ 2 }=0$$

This is quadratic in $$x$$ and will have only one root

$$\Rightarrow { (-4a{ m }^{ 2 }) }^{ 2 }-4(m)(-4{ a }^{ 2 })=0\\ 16{ a }^{ 2 }{ m }^{ 4 }+16{ a }^{ 2 }m=0\\ \Rightarrow m=-1$$

Point of contact to $$(i)$$ is $$\left( \dfrac { a }{ { m }^{ 2 } } ,\dfrac { 2a }{ m }  \right) $$

$$\Rightarrow (a,-2a)$$

Point of contact to $$(ii)$$ is $$\left( \dfrac { 2a }{ m } ,\dfrac { a }{ { m }^{ 2 } }  \right) $$

$$\Rightarrow (-2a.a)$$

which are respective ends of their latusrectum 

Hence proved.

 


Mathematics

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