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Locus of Vertex of y=a³x²/3+a²x/2-2a | JEE Parabola Family

JEE Maths question with a full step-by-step solution.

Question
The locus of the vertex of the family of parabolas y=a3x23+a2x2−2ay = \dfrac{a^3 x^2}{3} + \dfrac{a^2 x}{2} - 2a, where aa is a parameter, is:
Axy=10564xy = \dfrac{105}{64}correct
Bxy=34xy = \dfrac{3}{4}
Cxy=3516xy = \dfrac{35}{16}
Dxy=64105xy = \dfrac{64}{105}
Solution
Step 1: Factor out a33\dfrac{a^3}{3} and complete the square.
y=a33[(x+34a)2−10516a2]⇒y+105a48=a33(x+34a)2.y = \frac{a^3}{3}\left[\left(x + \frac{3}{4a}\right)^2 - \frac{105}{16a^2}\right] \Rightarrow y + \frac{105a}{48} = \frac{a^3}{3}\left(x + \frac{3}{4a}\right)^2.
 Step 2: The vertex is x=−34a, y=−105a48x = -\dfrac{3}{4a},\ y = -\dfrac{105a}{48}, so
xy=(−34a)(−105a48)=10564.xy = \left(-\frac{3}{4a}\right)\left(-\frac{105a}{48}\right) = \frac{105}{64}.
 Correct answer: (1)
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