If l is parallel to the line 2x-y + 4 = 0, then the equation of L is

If l is parallel to the line 2x-y + 4 = 0, then the equation of L is

Let L: 2x-y + k = 0
∵ l tangent to parabola
By eliminating y from y = x ^ 2 and 2x-y + k = 0, we get that:
x^2-2x-k=0
∴Δ=2^2-4*(-k)=0
k=-1
∴l：2x-y-1=0

How to find the tangent equation of a parabola with derivative

The tangent equation of a point (- 1,0) on the parabola y = x & sup2; + X is solved. Let the tangent equation be y-0 = k [x - (- 1)], that is, y = KX + K. substituting it into the parabola equation, we can get the following result: (x0dkx + k = x & sup2; + x-k) = (1-k) & sup2; + 4K = (1 + k) & sup2;; x0d is tangent, that is, there is only one intersection point, that is, the above equation has two equal real roots, (x0d △ = 0, the solution is k = - 1, so the tangent equation is y = - X-1

Let the distance from a point P to the straight line x + 2 = 0 on the parabola y2 = 4x be 5, then the distance from the point P to the focus F of the parabola is 5______ ．

The Quasilinear of parabola y2 = 4x is x = - 1, the distance between point P and straight line x + 2 = 0 is 5, and the distance between point P and quasilinear x = - 1 is 5-1 = 4. According to the definition of parabola, the distance between point P and the focus of parabola is 4, so the answer is: 4

It is known that P and Q are two points on the parabola x2 = 2Y. The abscissa of points P and Q are 4, - 2 respectively. If the two tangents cross point a, the ordinate of point a is ()
A. 1B. 3C. -4D. -8

∵ P, q are two points on the parabola x2 = 2Y, the abscissa of points P and Q are 4, - 2, ∵ P (4,8), q (- 2,2), ∵ x2 = 2Y, ∵ y = 12x2, ∵ y ′ = x, ∵ tangent equation AP, the slope of AQ, KAP = 4, KAQ = - 2, ∵ tangent equation AP is Y-8 = 4 (x-4), i.e., y = 4x-8, the abscissa of tangent equation AQ is Y-2 = - 2 (x + 2), i.e., y = - 2x-2, so that y = 4x-8y = - 2x-2, ∵ x = 1y = - 4, the ordinate of point a is -4. Select: C