The locus of the vertex of the family of parabolas y = 3 a 3 x 2 + 2 a 2 x − 2 a (a is parameter) is

For y = 3 a 3 x 2 + 2 a 2 x − 2 a , the vertex has x = − 2 a 3 /3 a 2 /2 = − 4 a 3 . Then y = 3 a 3 ⋅ 16 a 2 9 − 2 a 2 ⋅ 4 a 3 − 2 a = 16 3 a − 8 3 a − 2 a = − 16 35 a . Hence x y = ( − 4 a 3 ) ( − 16 35 a ) = 64 105 , so the locus of the vertex is x y = 64 105 .

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The locus of the vertex of the family of parabolas $y = \frac{a ^{3} x ^{2}}{3} + \frac{a ^{2} x}{2} - 2 a$ (a is parameter) is

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Solution:

For $y = \frac{a ^{3}}{3} x^{2} + \frac{a ^{2}}{2} x - 2 a$ , the vertex has $x = - \frac{a ^{2} /2}{2 a ^{3} /3} = - \frac{3}{4 a}$ .

Then $y = \frac{a ^{3}}{3} \cdot \frac{9}{16 a ^{2}} - \frac{a ^{2}}{2} \cdot \frac{3}{4 a} - 2 a = \frac{3 a}{16} - \frac{3 a}{8} - 2 a = - \frac{35 a}{16}$ .

Hence $x y = (- \frac{3}{4 a}) (- \frac{35 a}{16}) = \frac{105}{64}$ , so the locus of the vertex is $x y = \frac{105}{64}$ .

Stream:JEESubject:MathematicsTopic:ParabolaSubtopic:General

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