161 - Mathematics Questions Answers
![blush blush](http://www.iitbrain.com/plugin/ckeditor/plugins/smiley/images/embaressed_smile.gif)
Consider a branch of the hyperbola x²-2y²-2√2x-4√2y-6=0, with vertex at the point A. Let B be one of the end points of its latus rectum .If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is ??????![smiley smiley](http://www.iitbrain.com/plugin/ckeditor/plugins/smiley/images/regular_smile.gif)
Asked By: AMIT DAS
is this question helpfull: 6 0
submit your answer
Joshi sir comment
x²-2y²-2√2x-4√2y-6=0
Arrange this equation in form of standard hyperbola as
(x-√2)2/4 - (y+√2)2/2 = 1
so X = x-√2, Y = y+√2
vertex coordinate : X = 0 and Y = 0 so x = √2, y = -√2
similarly focus : X = ae, Y = 0, here a = 2, b = √2 and b2 = a2(1-e2)
and end point of latus rectum : X = ae, Y = b2/a
solve the area and get the answer