Q25 / 53 · NTA Abhyas (JEE Main) 2020 · Mathematics · Ellipse

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

Solution

Given equation is $\frac{a^{2}}{4} + b^{2} = 1$

Putting $a = 2 \cos θ, b = \sin θ$ , we get,

$2 a + 3 b$ $= 2 (2 \cos θ) + 3 (\sin θ) = 4 \cos θ + 3 \sin θ$
$\Rightarrow$ Minimum value $= - \sqrt{4^{2} + 3^{2}} = - 5$

Hence, the correct answer is option: 'B'.