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$Points\quad of\quad intersection\quad will\quad be\quad given\quad by\\ { x }^{ { 2 } }(\frac { 1 }{ { a }^{ 2 } } -\frac { { 1 } }{ \alpha ^{ 2 } } )+{ y }^{ { 2 } }(\frac { 1 }{ { b }^{ 2 } } -\frac { { 1 } }{ { \beta }^{ 2 } } )=0\\This\quad equation\quad represents\quad a\quad pair\quad of\quad straight\quad lines\quad through\\the\quad origin.\\Let\quad y=mx\quad satisfy\quad this\quad equation.\\ \Rightarrow (\frac { 1 }{ { a }^{ 2 } } -\frac { { 1 } }{ \alpha ^{ 2 } } )+{ m }^{ 2 }(\frac { 1 }{ { b }^{ 2 } } -\frac { { 1 } }{ { \beta }^{ 2 } } )=0\\ Using\quad properties\quad of\quad conjugate\quad diameters,\quad \\{ m }_{ 1 }{ m }_{ 2 }=\frac { (\frac { 1 }{ { a }^{ 2 } } -\frac { { 1 } }{ \alpha ^{ 2 } } ) }{ (\frac { 1 }{ { b }^{ 2 } } -\frac { { 1 } }{ { \beta }^{ 2 } } ) } =-\frac { { b }^{ 2 } }{ a^{ 2 } }\\ \Rightarrow 1-\frac { { a }^{ 2 } }{ { \alpha }^{ 2 } } =-1+\frac { b^{ 2 } }{ { \beta }^{ 2 } }\\\therefore \quad \frac { { a }^{ 2 } }{ { \alpha }^{ 2 } } +\frac { b^{ 2 } }{ { \beta }^{ 2 } } =\boxed{2}.$

Let the one of the points of intersection of the two ellipses be $(a\cos \phi, b\sin \phi)$ . As the points of are the ends of conjugate diameters, another point of intersection is given by $(-a\sin \phi, b\cos \phi)$ .

Now, substituting these points into the equation of the second ellipse, we get $\frac{a^2}{\alpha^2} \cos^2\phi+\frac{b^2}{\beta^2} \sin^2\phi=\frac{a^2}{\alpha^2} \sin^2\phi+\frac{b^2}{\beta^2} \cos^2\phi=1$

Therefore, $\frac{a^2}{\alpha^2}+\frac{b^2}{\beta^2}=2$