The tangent of parabola y = x2 at point P is perpendicular to the straight line 2x-6y + 5 = 0. The coordinate and tangent equation of point P are obtained Let's talk more about how the slope comes from

The tangent of parabola y = x2 at point P is perpendicular to the straight line 2x-6y + 5 = 0. The coordinate and tangent equation of point P are obtained Let's talk more about how the slope comes from

y=x²
Derivation
y'=2x
Straight line 2x-6y + 5 = 0, slope is 1 / 3
So the slope of the tangent is - 3
When 2x = - 3, x = - 3 / 2
Y = x & # 178;
Y = 9 / 4
therefore
The tangent coordinates are (- 3 / 2,9 / 4)
Tangent to
y-9/4=-3(x+3/2)
y=-3x-9/2+9/4
y=-3x-9/4

Elden Ring Premiere

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