WBJEE · Maths · Parabola
The locus of the vertices of the family of parabolas \(6 y=2 a^{3} x^{2}+3 a^{2} x-12 a\) is
- A \(x y=\frac{105}{64}\)
- B \(x y=\frac{64}{105}\)
- C \(x y=\frac{35}{16}\)
- D \(x y=\frac{16}{35}\)
Answer & Solution
Correct Answer
(A) \(x y=\frac{105}{64}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} 6 y+12 a &=2 a\left(a^{2} x^{2}+\frac{3}{2} a x\right) \\ &=2 a\left(a^{2} x^{2}+2 \cdot \frac{3}{4} a x+\frac{9}{16}\right)-\frac{9}{8} a \\ \Rightarrow 6 y+\frac{105}{8} a &=2 a\left(a x+\frac{3}{4}\right)^{2} \end{aligned}\) Let Vertices be (…
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