IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

sam_lavoisier

A coplanar beam of light emerging from a point source have equation

x-y+2(1+

) = 0 ,

R. The raya of the beam strike an elliptical surface and get reflected. The reflected rays from another convergent beam having equation

x-y+2(1+

) = 0

R . Further it is found that the foot of the perpendicular from the point (2,2) upon the tangent to the ellipse lies on the circle x² + y² - 4y - 5=0.

Find the eccentricity of the ellipse.And the area of the largest triangle that an incident ray and the corresponding reflected ray can enclose with the axis of the ellipse.

aman23iit

hi friend its simple

see first the eccentricity

the given equation is not of ellise as h²-ab<0

pzzz check the Q

sudeep.kumar

Lx - y + 2(1+L) = 0 can be written as

L(x+2) +(2-y) = 0

So, all the rays of the beam, which is represented by the above general formula passes thru the intersection of x+2=0 and 2-y=0... which is (-2,2)

So far, it is all right. But after that, you have given the same equation for the family of reflected beam... just changing the parameter.. and this means that all the reflected rays will also pass through the same point (-2,2)...

But this is not possible in the ellipthical mirror.. as there is no 'centre of curvature' in ellipse.. where all the normals concide...

hence... recheck the question... there is some missprint in it.. because what i m guessin is that the eqn. of reflected rays should be given in such a way that we get two focii of the ellipse (because in ellipse, all rays originating from focus converse on the other focus after reflection)...and then it wud be all cordinate maths to get 'e'.

amar.gupta

I also agree with Mr. Sudeep .

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