MCQ
If the equation $\frac{\lambda(\text{x}+1)^2}{3}+\frac{(\text{y}+2)^2}{4}=1$ represents a circle then $\lambda$:
- A$1$
- B$\frac{3}{4}$
- C$0$
- D$-\frac{3}{4}$
Solution:
$\frac{\lambda(\text{x}+1)^2}{3}+\frac{(\text{y}+2)^2}{4}=1$
for a circle of a $(\text{x}-\alpha)^2+6 (\text{y}-\beta)^2=1$ then (a = 6)
$\frac{\lambda}{3}=\frac{1}{4}$
$\lambda=\frac{3}{4}$
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