Pass a little P 1 1 and with hyperbola x 1 4 y 1

Pure on the ground floor.

It is true that there are four, but not all of them: two are tangents, and two are just intersections and not tangents. And these four common points are all on the right branch of the hyperbola.

There are two ways to know, the first is the drawing method

First you need to draw a basic image of the hyperbola, including asymptotics and vertices, and then find that the point p is on the outside of the hyperbola. When the straight line passing through point p is parallel to one of the two asymptotic lines, there is only one intersection point with the right branch of the hyperbola, and no intersection point with the left branch (as can be seen from the figure, in the first branch.

The second and third quadrants are always at a certain distance from the asymptotic line, and since the asymptote line and the hyperbola do not intersect, the straight line has no intersection point).

Then there can be two tangents to the right branch of the hyperbola that cross the p point, and these two tangents are in the first.

The second and third quadrants are farther and farther away from the asymptote line, so there is no intersection with the left branch of the hyperbola. Of course, there are also two tangents to the left branch of the hyperbola at the point p, but these two tangents are at the first.

The first and fourth quadrants all have intersections with the right branch of the hyperbola.

The second is the analytic method: let the linear equation be y=k(x-1)+1, and eliminate y with the hyperbola

The resulting equation (k -4) x -2k (k-1) x + k -2k + 5 = 0

When k -4 = 0, that is, k = 2, the equation becomes a univariate one-dimensional equation and must have a unique solution;

When k -4≠0, =[2k(k-1) -4(k -4)(k -2k+5)=16(5-2k)=0, the solution is k=

In addition, when k is infinite, i.e., when the line is perpendicular to the x-axis, the linear equation is x=1, and there is a unique solution to the hyperbola. So there are a total of four straight lines with a single common point with a hyperbola.

It's 4 indeed.

You can understand this qualitatively by drawing the following diagram, and there is no need for quantitative calculation.

According to the hyperbola formula x 2-(1 4)y 2=1, we can know that this hyperbola is symmetrical with respect to the y-axis, and passes through (1,0) and (-1,0), then there should be four straight lines tangent to the hyperbola through (1,1), the tangent points are in four quadrants, of which the first and second quadrants are at infinity, and the third and fourth quadrants are probably closer to (1,0) and (-1,0), and the four tangent points are symmetrical with respect to the y-axis.

You can draw a simple diagram and you will understand, hehe.

The hyperbola x -y 4=1, the straight line l of the crossing point p(1,1) has only one common point with the hyperbola, find the equation of the straight line l.

Solution: When the slope of the straight line does not exist, the straight line is x=1, and the straight line x=1 has only one common point with the hyperbola, which is in line with the topic.

When the slope of the straight line exists, let the linear equation of the straight line l be: y-1=k(x-1), the simultaneous linear equation and the hyperbolic equation, and obtain:

x^2+2k(k-1)x-k^2+2k-5=0

1.When 4-k 2 = 0, i.e. k = 2, the equation is a one-dimensional equation with only one solution.

There is only one intersection point, and the equation for the straight line is y-1=2 (x-1) or y-1=-2 (x-1).

i.e. y=2x-1, or y=-2x+3

2.When 4-k 2≠0:

4k 2(k-1) 2-4(4-k 2) (k 2+2k-5)=0 with an intersection.

Then: k=5 4, and the linear equation is y-1=5 4*(x-1), i.e., 5x-4y-1=0

In summary: l has 4 articles, namely: x=1 or y=2x-1, or y=-2x+3 or 5x-4y-1=0

The point p(1,1) is not on the hyperbola.

a = 1a = 1 and the real axis is 1

So tangent to hyperbola x=1 is a straight line.

The left focus is set to a, and the right focus is b

So|pa|-|pb|=2a=6

The two circles of m and n are exactly the two foci of the hyperbola.

i.e., m is on a, n is on b, a radius is 2, b radius is 1, where |pm|≤|pa|-|am|If and only if M is true on the PA.

pn|≥|pb|+|bn|If and only if N is established on the PB extension.

So |pm|-|pn|≤|pa|-|am|-|pb|-|bn|=(pa|-|pb|)-am|-|bn|

So min(|pm|-|pn|) = 3 if and only if M is on the PA and N is true on the PB extension.

Draw their images first. When three points are in the line. p has a minimum value of 6 when it is in the middle

P is known to be hyperbola.

x 25-y 9=1, a(5,0),b(-5,0), then kpa kpb is ?

Solution: Let p(m, n) be a point on the hyperbola x 25-y segment Luyu 9 =1, then m 25-n 9 =1

Get n =9 (m 25 1).

a(5, 0), b(5, 0).

kpa kpb =n m-5 n m+5 =n m -25 = 9 (m grip rock 25-1) m -25 = 9 25

i.e. k pa kpb up-and-out = 9 25

Note]: General conclusion: k pa kpb = hyperbolic focus on the x-axis, a, b are vertices).

y-1=k(x-1)

y=kx+(1-k)

Substituting 2x 2-y 2=2

2-k2) x 2-2k(1-k)x-(1-k) 2-2=0p is the midpoint, then (x1+x2) 2=1

So x1+x2=2k(1-k) (2-k2)=2k-k2=2-k2

k=2 so yes.

The asymptote equation is y=x2, i.e., x2y=0, the point p coordinates are (m,n), and m4 n =1, so m-4n=4

So the distance from p to two straight lines d1=|m+2n|/√5，d2=|m-2n|/√5

So d1d2=|m²-4n²|5=4 5, which is a constant.

pa|²=(m-3)²+n²=(m-3)²+m²/4-1=5m²/4-6m+8=5/4(m-12/5)²-104/5

Because m 2 or m -2, the minimum value is obtained at -2 or 12 5, and because m = 12 5, the value is -104 5<0, so it is discarded.

So when m=-2, |pa|=25, so the minimum value of 丨pa丨 is 5

Images can be used to solve qualitatively.

1) The straight line is tangent to the hyperbola, and there are two.

2) The straight line is parallel to the asymptote line, and there are two.

Therefore, there are 4 straight lines with only one intersection point between the point p(1,1) and the hyperbola x 2 9-y 2 16=1.

I think the answer on the second floor is correct, and the intersection points are all on the right branch of the hyperbola.

The point p(1,1) is on the outside of the curve of the double pulse family and is not on the proximal line of the gradual.

Passing the point p(1,1) can make two parallel lines and hyperbolas with an intersection point.

Three of a kind. The first is a straight line parallel to the y-axis through point a, tangent to the right branch hyperbola and (3,0). The second and third are straight lines that cross point a, respectively, parallel to the progressive line of this hyperbola (which has two progressive lines).

Two. The p-point is on the asymptote.