Given that curve C: y = lnx-4x and line x = 1 intersect at a point P, then the tangent equation of curve C at point P is______ ．

It is known that y ′ = 1 x-4, so when x = 1, there is y ′ = - 3, that is, the slope of the tangent passing through point P K = - 3, and y = ln 1-4 = - 4, so the tangent point P (1, - 4), so the tangent equation at point P is y + 4 = - 3 (x-1), that is, 3x + y + 1 = 0. So the answer is 3x + y + 1 = 0

What is the tangent equation perpendicular to x + y = 1 on the curve y = INX?

y=-x+1
Perpendicular to him, the slope is 1
y=lmx
y'=1/x
Then the slope k = y '= 1
x=1
y=lnx=0
Tangent point (1,0)
So it's x-y-1 = 0

Given the curve y = x ^ 3 + 3x ^ 2-5, find the tangent equation and normal equation of (1) passing through point (- 1, - 3)

If y = 3x ^ 2 + 6x, the derivative at (- 1, - 3) is - 3
That is, the tangent slope at this point is - 3
Let the tangent equation be y = - 3x + B
Substituting (- 1, - 3) points, we can get b = - 6, so the tangent equation is y = - 3x-6
The normal is perpendicular to the tangent and the slope should be 1 / 3
So y = x / 3 + B, substituting (- 1, - 3) points, we get b = - 8 / 3, so the normal equation is 3Y = X-8

Find a point M0 on the curve y = root x, make the tangent passing through the point M0 parallel to the straight line x-2y + 5 = 0, and find the tangent equation and normal equation passing through the point M0 (detailed explanation)

The tangent passing through the point M0 is parallel to the straight line x-2y + 5 = 0
Then the tangent slope k = 1 / 2
y'=1/(2√x)
Then the tangent property k = 1 / (2 √ M0 = 1 / 2)
The solution is M0 = 1
So the tangent point is (1,1)
The tangent equation is y = (1 / 2) (x-1) + 1, that is y = (1 / 2) x + 1 / 2
Normal slope k '= - 1 / k = - 2
The normal equation is y = - 2 (x-1) + 1, that is y = - 2x + 3

Given that curve C: y = lnx-4x and line x = 1 intersect at a point P, then the tangent equation of curve C at point P is______ ．