Question

The locus of the vertices of the family of parabolas $straight y space equals fraction numerator straight a cubed straight x squared over denominator 3 end fraction space plus fraction numerator straight a squared straight x over denominator 2 end fraction minus 2 straight a space is$

$xy space equals space 105 over 64$
$xy space equals space 5 over 4$
$xy space equals space 35 over 16$
$xy space equals space 64 over 105$

Solution

xy space equals space 105 over 64

Parabola colon space straight y space equals fraction numerator straight a cubed straight x squared over denominator 3 end fraction plus fraction numerator straight a squared straight x over denominator 2 end fraction minus 2 straight a Vertex colon thin space left parenthesis straight alpha comma space straight beta right parenthesis straight alpha space equals space fraction numerator negative straight a squared divided by 2 over denominator 2 straight a cubed divided by 3 end fraction space equals space minus space fraction numerator 3 over denominator 4 straight a end fraction comma straight beta space equals space fraction numerator open parentheses begin display style straight a to the power of 4 over 4 end style plus 4. begin display style straight a cubed over 3 end style.2 straight a close parentheses over denominator 4 begin display style straight a cubed over 3 end style end fraction space space equals space fraction numerator open parentheses negative begin display style 1 fourth end style plus begin display style 8 over 3 end style close parentheses straight a to the power of 4 over denominator begin display style 4 over 3 end style straight a cubed end fraction αβ space equals space minus space fraction numerator 3 over denominator 4 straight a end fraction open parentheses negative 35 over 16 close parentheses straight a space equals space 105 over 64

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