 Hello and welcome to the session. Let's work out the following question. It says a point on the hypotenuse of a triangle Is that a distance a and b from the sides of the triangle? Show that the minimum length of the hypotenuse is a to the power 2 by 3 plus b to the power 2 by 3 whole to the bar 3 by 2 so let's now move on to the solution and let be be the point on the hypotenuse a C of Right-angled triangle a b c such that P L is perpendicular to a b and P m is perpendicular to BC and let Angle a C B be equal to theta Now since P L is parallel to BC and a C is a transversal therefore angle APL Will also be equal to theta Now let Length of the hypotenuse B L Now L is equal to AP plus PC now since angle a P L is theta And P L is a therefore a P is equal to a secant theta Similarly in right triangle P mc angle PC m is theta and this Distance that is P m is b therefore PC will be equal to b cosic theta right Now we'll find the L y d theta Derivative of secant theta is secant theta into tan theta and derivative of cosecant theta is minus cosecant theta cot theta and here theta lies between 0 to pi by 2 now for maximum or Minimum we need to put dl by d theta equal to 0 so this implies a secant theta into tan theta minus b cosecant theta Into cot theta equal to 0 So this implies a into 1 upon cos theta sin theta is tan theta sin theta upon cos theta Minus b into cosec theta that is 1 upon sin theta Into cot theta that is cos theta upon sin theta equal to 0 So this implies a sin theta upon cos square theta minus b cos theta Upon sin square theta equal to 0 And this implies a sin theta upon cos square theta is Equal to b cos theta upon sin square theta and this implies a sin cube theta is equal to b cos cube theta and this implies tan cube theta is equal to b by a and this implies tan theta is equal to b by a to the power 1 by 3 Now we need to find the second-order derivative dl by d theta is a sec theta tan theta minus b cosec theta into cot theta So we need to apply the product rule to the first term and the second term so it is a into secant theta into derivative of tan theta that is secant square theta plus tan theta into derivative of secant theta that is secant theta into tan theta minus b into derivative of cosecant theta into cot theta now derivative of cosecant theta is minus cosecant theta into cot theta into cot theta plus plus cosecant theta as it is into derivative of cot theta that is minus cosecant square theta. Now further this implies d square l upon d theta square is equal to a into secant cube theta plus tan square theta into secant theta taking minus common from this expression it becomes plus so we have b into cosec theta into cot square theta plus cosecant cube theta. Now again taking secant theta common we have a into secant theta into secant square theta plus tan square theta plus b into taken cosecant theta common we have b into cosecant theta into cot square theta plus cosecant square theta. Now we have to find the minimum length now for theta lying between 0 to pi by 2 secant theta and cosecant theta is greater than 0 and also the expression inside the bracket in both the brackets is greater than 0 since they are perfect square being a square therefore the whole expression is greater than 0 therefore d square l by d theta square is greater than 0 for or when tan theta is equal to b by a to the power 1 by 3. Now we have to find the minimum length now l is equal to a secant theta plus b cosecant theta now secant theta can be written as square root of 1 plus tan square theta plus b into cosecant theta can be written as square root of 1 plus cot square theta now tan theta is b by a to the power 1 by 3 so this becomes b by a to the power 2 by 3 plus b into square root of 1 plus cot square theta now tan theta is b by a to the power 1 by 3 so cot theta will be a by b to the power 1 by 3 and cot square theta will become a by b to the power 2 by 3. This is further equal to a into square root of a to the power 2 by 3 taking LCM plus b to the power 2 by 3 upon a to the power 1 by 3 cancelling square with the square root. We get this plus b into again taking LCM we have b to the power 2 by 3 plus a to the power 2 by 3 upon b to the power 1 by 3 thus further equal to a to the power 2 by 3 into square root of a to the power 2 by 3 plus b to the power 2 by 3 plus b to the power 2 by 3 into square root of b to the power 2 by 3 plus a to the power 2 by 3 taking square root of a to the power 2 by 3 plus b to the power 2 by 3 common we have square root of a to the power 2 by 3 plus b to the power 2 by 3 into a to the power 2 by 3 plus b to the power 2 by 3. Now square root of a to the power 2 by 3 plus b to the power 2 by 3 into a to the power 2 by 3 plus b to the power 2 by 3 is equal to a to the power 2 by 3 plus b to the power 2 by 3 to the power 3 by 2. Hence the minimum length of the hypotenuse is a to the power 2 by 3 plus b to the power 2 by 3 whole to the power 3 by 2 and that is what we had to prove hence proved. So this completes the question and the session. Bye for now. Take care. Have a good day.