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Minimum of (r+5-4cos)^2+(r-3sin)^2 Expression | JEE

JEE Maths question with a full step-by-step solution.

Question

The minimum value of

\left\{(r+5-4|\cos\theta|)^2+(r-3|\sin\theta|)^2\right\}

\forall\,r,\theta\in\mathbb{R}

, is:

0

correct

2

3

Dnone of these

Solution

Step 1: Read the expression as a squared distance. The point

(r+5,\ r)

lies on the line

y=x-5

, and the point

(4|\cos\theta|,\ 3|\sin\theta|)

lies on the ellipse

\dfrac{x^2}{16}+\dfrac{y^2}{9}=1

. The bracket is the squared distance between these two points. Step 2: So the minimum equals the squared shortest distance between the line

y=x-5

and the ellipse

\dfrac{x^2}{16}+\dfrac{y^2}{9}=1

. Step 3: The tangents to the ellipse with slope

1

are

y=x\pm\sqrt{16\cdot1+9}=x\pm5

. Thus

y=x-5

is itself a tangent to the ellipse, so the shortest distance is

0

\therefore

the minimum value is

0

. Correct answer: (1)

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