Conic Curve Puzzle, Conic Curve Problem?

Because the straight line y x a 1 is parallel to the hyperbolic x 2 a y 2 asymptote line y x 2, laugh at the brigade.

The straight line touches the stool y x-1 and intersects the right branch of the hyperbola at a point (the relationship with the left branch is separate). Big search.

That is to say: the straight line y x 1 and the hyperbola x 2 2 y 2 2 1 are intersecting, but there is only one intersection point!

y=x-1 is substituted for x 2-y 2=1, and x -(x-1) is noisy=2

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

Therefore, there is only one point of change in the sentence of lack of suspicion.

1.The coordinates of the setting point a b are (a, a2 4) (b, b2 4) respectively

Then the tangent equation for a b is ma:y=ax 2- a2 4 mb:y=bx 2- b2 4 (this step is relatively simple, you can calculate it yourself).

Then the intersection point of ma and mb is m((a+b) 2,ab 4), and ab 4 can be obtained

The straight line ab intersects the y-axis at the point (0,-ab 4), that is, the straight line ab passes the point (0,m).

2.From the above, the slopes of MA MB AB are A2, B2, (A+B) 4, respectively

First, the possibility that AMB is a right angle i.e., MA is perpendicular to MB and the slope is multiplied by -1

If we get the equation a 2*b 2=-1 then m=1, where amb is a right angle.

Then discuss the possibility that ABM is a right angle, i.e., AB is perpendicular to MB and the slope is multiplied by -1

If we get the equation (a+b) 2*b 2=-1 then ab+b2=-4, then b2=0 is not true i.e. abm is not a right angle.

In the same way, a mab cannot be a right angle.

So m=1, amb is at a right angle ab=-4

That is, as long as the points on the line l:y=-1 are satisfied, the mab is a right triangle.

[Note: Draw a more standardized diagram first, so that the numbers can be combined.] Solution:

You can set the right focus point to F2 and connect PF2, OM, and OTBy the hyperbolic equation (x 4) - (y 9) = 1It can be seen that a = 2, b = 3, c = 13

And|pf|-|pf2|=2a=4.===>|pf|=4+|pf2|.∴fm|=|pf|/2=2+(|pf2|/2).

a) In oft, it is easy to know, ot ft, of=13, ot=2. From the Pythagorean theoremft|=3, (b) In PFF2, OF=OF2, FM=MP, OM is the median line, let OM=X, then PF2=2X∴fm=2+(|pf2|2) = 2 + In summary, |om|=x,|tm|=x-1.

mo|-|mt|=x-(x-1)=1.i.e. |mo|-|mt|=1.

c a = 6 3, b 2 = a 2 - c 2 = 1 3a 2 ellipse c: x a + 3y a = 1

x^2+3y^2=a²

ab:x=y+c, substitution x 2+3y, 2=a (y+c), 2+3y, 2=a2

4y^2+2cy+c^2-a^2=0

Let a(x1,y1),b(x2,y2)n(x0,y0)2y0=y1+y2=-c/2,y1y2=(c^2-a^2)/4y0=-c/4,x0=3/4c

kon=y0/x0=-1/3

m(x,y)

om=moa+nob

mx1+nx2,my1+ny2)

mx1+nx2)^2+3(my1+ny2)^2-a²=0∴m^2(x1²+3y1²)+n^2(x2²+3y^2)+2mnx1x2+6mny1y2-a^2=0

m^2a^2 +n^2a^2+2mn(x1x2+3y1y2)-a^2=0

3y1y2=-3a^2/4,x1x2=(y1+c)(y2+c)=y1y2+c(y1+y2)+c^2

x1x2+3y1y2=4y1y2+c(y1+y2)+1= 4(c^2-a^2)/4+(-c^2/4)+c^2=-a^2/3-c^2/6+c^2=0

m^2a^2 +n^2a^2=a^2

m + n =1, m = cos , n = sin i.e. the total existence of the angle r makes the equation:

The vector om=cos vector oa+sin vector ob is established.

Content from the user: Leave a light to rotate.

tan130

tan（-130）

tan（-130+180）

tan50 for hyperbolas with the focus on the x-axis:

Asymptote slope = b a or -b a

Eccentricity = Ca

Because the square of a + the square of b = 1;

Simultaneous three formulas can wait until the eccentricity.

Analysis: B a is obtained from the known

tan50°, turn into a chord function, and then square the two sides to find the eccentricity of c