1.

Find the equation of the normal to curve `x^2=4y`which passes through the point (1, 2).

Answer» `h^2=4k`
slope of normal`=-1/(dy/dx)=-2/h`
equation of normal`(y-k)=-2/h(x-h)`
`k=2+2/h(1-h)`
`h^2/4=2+2/h(1-h)`
h=2
k=1
equation of line
(y-1)=-1(x-2)
x+y=3


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