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Grade 12th passPhysical Chemistry

if the curvees x^2/R+y^2/4=1 and y^2=16x intersect at right angles then value of ‘R’

Profile image of kalyan
6 Years agoGrade 12th pass
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1 Answer

Profile image of Vikas TU
6 Years ago
Let Both cut at angle (x1,y1)
x^2/R+y^2/4=1
so , on differenciating 
2x / R + 2y*y'/4 = 0
y' = -x/R *4/y 
Diiferenciating another curve 
2y *y' = 16 
y' = 8/y 
Both solpe is perpendicular 
-x1/R * 4/y1 * 8/y1  = -1 
y1 ^2 =16x1 
Put this value 
32 /16R = 1 
R = 2