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

Vikas TU
8825 Points
```							Let Both cut at angle (x1,y1)x^2/R+y^2/4=1so , on differenciating 2x / R + 2y*y'/4 = 0y' = -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
```
one month ago
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• 141 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions