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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