Let $P (h, k)$ be a fixed point where $h > 0, k > 0$ . A straight lien through $P$ cuts the coordinate axes $A$ and $B$ . The minimum area of the triangle $O A B$ is

h k

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\frac{h^{2} + k^{2}}{2}

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2 h k

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4 h k

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Solution

The correct option is B $2 h k$
Let the slope is m. then the equation of the line is y-k=m(x-h) It cuts the co ordinate axix at the points

A={(k-mh), 0} and B= $0, (h - m / k)$ hence the area is $Δ = 1 / 2 (k - m h) (h - m / k)$ then applying min condition for the variable m we get the min area is $= 2 h k$ .

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Let P(h,k) be a fixed point where h>0,k>0. A straight lien through P cuts the coordinate axes A and B. The minimum area of the triangle OAB is

The correct option is B 2hkLet the slope is m. then the equation of the line is y-k=m(x-h) It cuts the co ordinate axix at the points A={(k-mh), 0} and B=0,(h−m/k) hence the area is Δ=1/2(k−mh)(h−m/k) then applying min condition for the variable m we get the min area is =2hk.

Let $P (h, k)$ be a fixed point where $h > 0, k > 0$ . A straight lien through $P$ cuts the coordinate axes $A$ and $B$ . The minimum area of the triangle $O A B$ is