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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If a, b ∈ R satisfy the equation a^ 2 + 4b^ 2 - 4 = 0, then the...

If a,b∈Ra, b \in R satisfy the equation a2+4b2−4=0,a^{2} + 4b^{2} - 4 = 0, then the minimum value of (2a+3b)\left(2a + 3b\right) will be

Options

  1. A−4- 4
  2. B−5- 5
  3. C−6- 6
  4. D−10- 10

Correct answer

B. −5- 5

Step-by-step solution

Given equation isa24+b2=1\dfrac{a2}{4} + b^{2} = 1

Putting a=2cosθ,b=sinθa = 2cos \theta, b = sin \theta, we get,

2a+3b=2(2cosθ)+3(sinθ)=4cosθ+3sinθ2a + 3b= 2\left(2cos \theta\right) + 3\left(sin \theta\right) = 4cos\theta + 3sin\theta
⇒⇒ Minimum value =−42+32=−5= - \sqrt{{4}^{2} + {3}^{2}} = - 5

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