Find the equation of the normal at the point on the curve x^2=4y which passes through the point (1,2).Also fine the equation of the corresponding tangent.
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The equation of the given curve is x2=4y ...1Differentiating 1 w.r.t x, we get2x=4dydx⇒dydx=2x4⇒dydx=x2Now, dydx1,2=12 So, the equation of tangent at 1,2 is given by:y-y1=dydx1,2x-x1⇒y-2=12x-1⇒2y-4=x-1⇒2y-x-4+1=0⇒2y-x-3=0Equation of normal at 1,2 is given by:y-y1=-1dydx1,2x-x1⇒y-2=-112x-1⇒y-2=-2x-1⇒y-2=-2x+2⇒2x+y-4=0
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The equation of the given curve is x2=4y ...1Differentiating 1 w.r.t x, we get2x=4dydx⇒dydx=2x4⇒dydx=x2Now, dydx1,2=12 So, the equation of tangent at 1,2 is given by:y-y1=dydx1,2x-x1⇒y-2=12x-1⇒2y-4=x-1⇒2y-x-4+1=0⇒2y-x-3=0Equation of normal at 1,2 is given by:y-y1=-1dydx1,2x-x1⇒y-2=-112x-1⇒y-2=-2x-1⇒y-2=-2x+2⇒2x+y-4=0
Regards
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