If the parabolic equation is: (Y &; K) 2 = 4C (X &; H), then the tangent equation passing through a point (x0, Y0) on the parabola is derivative

The derivation of X on both sides shows that the tangent slope of 2 (Y-K) y '= 4cy' = 2C / (Y-K) is 2C / (y0-k) (Y0 &; K) &# 178; = 4C (x0 &; H), (Y0 &; K) = 4C (x0 &; h) / (Y0 &; K), and the point oblique equation is y = Y0 + 2C / (y0-k) * (x-x0) = K + (y0-k) + 2C [(x

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