# find the equation of the normal to the curve y = root (2x^2 + 1) at the point (2,3) The slope of the tangent to a curve at any point is the value of the first derivative at that point.

Here, we have the curve y = sqrt ( 2x^2 +1)

y' = 4x * (1/2) (2x^2 + 1)^(-1/2)

=> y' = 2x/ sqrt (2x^2 + 1)

At the...

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The slope of the tangent to a curve at any point is the value of the first derivative at that point.

Here, we have the curve y = sqrt ( 2x^2 +1)

y' = 4x * (1/2) (2x^2 + 1)^(-1/2)

=> y' = 2x/ sqrt (2x^2 + 1)

At the point (2,3), y' = 2*2 / sqrt (2*4 + 1)

=> 4 / sqt 9

=> 4/3

As the slope of the tangent is 4/3, the slope of the normal which is perpendicular to the tangent is -3/4.

The equation of a line passing through (2,3) and with a slope -3/4 is

(y - 3)/ (x - 2) = -3/4

=> 4y - 12 = -3x + 6

=> 3x + 4y - 18 = 0

The equation of the normal is 3x + 4y - 18 = 0.

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