A variable line L passes through the point (3, 5) and intersects the positive coordinate axes at the points A and B. The minimum area of the triangle O A B , where O is the origin, is: [2024]

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Solution

Q.
A variable line $L$ passes through the point (3, 5) and intersects the positive coordinate axes at the points $A$ and $B$ . The minimum area of the triangle $O A B,$ where $O$ is the origin, is: [2024]

1 35

2 40

3 25

4 30

Ans.
(4)

Equation of line $A B : \frac{x}{a} + \frac{y}{b} = 1$

Since, it passes through (3, 5)

$∴$    $\frac{3}{a} + \frac{5}{b} = 1$

$\Rightarrow 3 b + 5 a = a b \Rightarrow 5 a - a b = - 3 b \Rightarrow a (5 - b) = - 3 b$

$\Rightarrow a = \frac{3 b}{b - 5}$

Area of triangle $=$ $| \frac{1}{2} \times a \times b |$ $= \frac{1}{2} \times \frac{3 b}{b - 5} \times b$

$\Rightarrow f (b) = \frac{3 b^{2}}{2 b - 10}$

$\Rightarrow f' (b) = 0 \Rightarrow (2 b - 10) 6 b - 2 (3 b^{2}) = 0$

$\Rightarrow 12 b^{2} - 60 b - 6 b^{2} = 0 \Rightarrow 6 b^{2} - 60 b = 0$

$\Rightarrow b^{2} - 10 b = 0 \Rightarrow b (b - 10) = 0$

   $b = 0 or b = 10$

So for minimum area, $b = 10$

$Hence, minimum area = \frac{3 b^{2}}{2 b - 10} = \frac{3 \times {(10)}^{2}}{2 \times 10 - 10}$

   $= \frac{300}{10} = 30 sq. units$

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