find equation of normal of curve 2y=x^2, which passes through point(2,1)
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given curve 2y=xsquare
diff w.r.t x
y=x square÷2
dy/dx=x
at 2,1 slope =2
normal eq=x+2y-4=0
diff w.r.t x
y=x square÷2
dy/dx=x
at 2,1 slope =2
normal eq=x+2y-4=0
Answered by
21
Answer:
wrong ans. ,plz correct it.
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