area

Question

The vertices of a triangle are (0, 0),(sec θ, tanθ) and (4, -5) (0<θ<&prod;/2). The minimum possible area is --------

Badiuddin askIITians.ismu Expert · Accepted Answer

Dear ANAM open determinant &Delta; =1/2| [-5sec&Theta; -4tan&Theta;]| &Delta; =1/2| [5sec&Theta; +4tan&Theta;]| clearly at &Theta; =0 both tan&Theta; and sec &Theta; is minimum in there given domain. so put &Theta;=0 &Delta; =5/2 Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin

Analytical Geometry> area...

The vertices of a triangle are (0, 0),(sec θ, tanθ) and (4, -5) (0<θ<∏/2). The minimum possible area is --------

ANAM V. RANGA REDDY , 16 Years ago

Badiuddin askIITians.ismu Expert

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Analytical Geometry> area...

The vertices of a triangle are (0, 0),(sec θ, tanθ) and (4, -5) (0<θ<∏/2). The minimum possible area is --------

ANAM V. RANGA REDDY , 16 Years ago

Badiuddin askIITians.ismu Expert

LIVE ONLINE CLASSES

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