Question

If the equation $\frac{\lambda (x+1{)}^2}{3}+\frac{(y+2{)}^2}{4}=1$ represents a circle then $\lambda =?$

• A. $1$
• B. $\frac{3}{4}$
• C. $0$
• D. $-\frac{3}{4}$

Answer 1

Answer: (B)

$\frac{3}{4}$

$\frac{\lambda (x+1{)}^2}{3}+\frac{(y+2{)}^2}{4}=1$

for a circle of $a(x-\alpha {)}^2+6(y-\beta {)}^2=1$ then $(a=6)$

hence

$\frac{\lambda }{3}=\frac{1}{4}$

$\lambda =\frac{3}{4}$