Wo equal parabolas have the same vertex and their axes are at right angles. Prove that their common tangent touches each at the end of their latus
Answers
Let the two parabolas be { y }^{ 2 }=4ax......(i)y
2
=4ax......(i) and { x }^{ 2 }=4ay......(ii)x
2
=4ay......(ii)
Equation of tangent to (i)(i) is
y=mx+\dfrac { a }{ m }y=mx+
m
a
It also touches (ii)(ii) so substituting yy in (ii)(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 }=0x
2
=4a(mx+
m
a
)
mx
2
=4am
2
x+4a
2
mx
2
−4am
2
x−4a
2
=0
This is quadratic in xx 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⇒(−4am
2
)
2
−4(m)(−4a
2
)=0
16a
2
m
4
+16a
2
m=0
⇒m=−1
Point of contact to (i)(i) is \left( \dfrac { a }{ { m }^{ 2 } } ,\dfrac { 2a }{ m } \right)(
m
2
a
,
m
2a
)
\Rightarrow (a,-2a)⇒(a,−2a)
Point of contact to (ii)(ii) is \left( \dfrac { 2a }{ m } ,\dfrac { a }{ { m }^{ 2 } } \right)(
m
2a
,
m
2
a
)
\Rightarrow (-2a.a)⇒(−2a.a)
which are respective ends of their latusrectum
Hence proved.