the equation of the normal to ellipse x^2+4y^2=16 at the end of the latus rec tumin the first quadrant is
(a) 2x+√3(y+3)=0 (b)2x=√3(y+3) (C) √3x=2(y+3) (d)none
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UP 0 DOWN 0 0 2
2 Answers
1x^2+4y^2=16
x^2/16+y^2/4=1
e^2=1-4/16 e=root3/2 ae=2root3
the y coordinate on the ellips in 1st quadrant is 1
find line passing tru (2root3,1)
only b satisfies..