Question

If $a,b\in R$ satisfy the equation $a^2+4b^2-4=0,$ then the minimum value of $\left(2a+3b\right)$ will be

• A. $-4$
• B. $-5$
• C. $-6$
• D. $-10$

Answer 1

Answer: (B)

Given equation is $\frac{a^2}{4}+b^2=1$
Putting $a=2\cos \theta ,b=\sin \theta$ , we get,
$2a+3b$ $=2\left(2\cos \theta \right)+3\left(\sin \theta \right)=4\cos \theta +3\sin \theta$
$\Rightarrow$ Minimum value $=-\sqrt{4^2+3^2}=-5$