High School Hyperbolic Questions 60, High School Hyperbolic Questions

The line l:y=kx+1 intersects with hyperbola c at 2x -y =1 at two points ab, find the range of k?

Is there a real number k such that a circle with the diameter of the line segment ab passes through the right focus f of the hyperbola to find the value of k?

Analysis: Straight line l: y=kx+1==>y 2=k 2x 2+2kx+1

Substituting hyperbolic c: 2x 2-y 2=1

, (2-k2)x 2-2kx-2=0

From Veda's theorem: x1+x2=2k (2-k 2), x1x2=-2 (2-k 2).

From the inscription, AFB is a right triangle, i.e. AF FB

f(0, √6/2)

k(af)=y1/(x1-√6/2), k(bf)=y2/(x2-√6/2)

y1y2/[(x1-√6/2)(x2-√6/2)]=[k^2x1x2+k(x1+x2)+1]/[x1x2-√6/2(x1+x2)+3/2]=-1

k^2+1)x1x2+(k-√6/2)(x1+x2)+5/2=0

2(k^2+1)/(2-k^2)+ 2k(k-√6/2)/(2-k^2)+5/2=0

2-√6k)/(2-k^2)+5/2=0

5k^2+2√6k-6=0

k1=-(6+ 6) 5,k2=(6- 6) 5 (undesirable).

There is a real number k=-(6+ 6) 5 such that a circle with the diameter of the line segment ab passes through the right focus f of the hyperbola

**Save first... Zoom in and see . See **, in more detail.

Solution: Composition: f2e pf1

Because, the distance from F2 to the straight line Pf1 is equal to the real long axis.

So, f2e=2a, because |pf2|=|f1f2|=2c

In the isosceles triangle F1F2P, because, F2E Pf1, PE=EF1=Pf1 2

In RT F1EF2, EF1 = under the root number [(F1F2) -F2E)] = under the root number [(2C) -2A)]=2b

So, pf1 = 2ef1 = 4b

From the definition and title of hyperbola: pf1-pf2=2a (there is a point p in the right branch of the hyperbola), i.e., 4b-2c=2a

So, c = 2b-a substitution, c = a +b (2b-a) = a +b

So, 3b -4ab = 0

So, b a = 4 3

So, the asymptote is wrought: y=(4 3)x and y= (4 3)x

The straight line l:y = root number 3 (x-2) and the hyperbolic x2 a2-y2 b2 = 1 intersect at two points ab, ab = root number 3, there is a line l about l'y=b ax symmetry line l2 is parallel to the x-axis.

Find the eccentricity and hyperbolic equations.

The simultaneous line y= 3 (x-2) with hyperbola x 2 a 2-y 2 b 2 = 1 yields:

b^2x^2-a^2y^2-a^2b^2=0

> b^2x^2-a^2*3(x-2)^2-a^2b^2=0

> b^2x^2-3a^2(x^2-4x+4)-a^2b^2=0

> (b^2-3a^2)x^2+12a^2x-(12a^2+a^2b^2)=0

> x1+x2=12a^2/(3a^2-b^2)，x1x2=(12a^2+a^2b^2)/(3a^2-b^2)

So: (x1-x2) 2=(x1+x2) 2-4x1x2

12a^2/(3a^2-b^2)]^2-4(12a^2+a^2b^2)/(3a^2-b^2)

144a^4-4*(12a^2+a^2b^2)*(3a^2-b^2)]/(3a^2-b^2)^2

4a^2b^2*(12-3a^2+b^2)/(3a^2-b^2)^2

Again, y1 = 3 (x1-2) and y2 = 3 (x2-2).

So, y1-y2 = 3 (x1-x2).

So, (y1-y2) 2=3(x1-x2) 2

Then, ab 2=(x1-x2) 2+(y1-y2) 2=4(x1-x2) 2=3

> 16a^2b^2*(12-3a^2+b^2)/(3a^2-b^2)^2=3………1）

The slope of the straight line l:y = 3 (x-2) is k1 = 3

Straight line l':The slope of y=(b a)x is k=b a

The slope of the straight line l2 is k2=0

Because the straight lines l1 and l2 are symmetrical with respect to l.

So: (k-k1) (1+kk1) = (k2-k) (1+kk2).

> (b/a-√3)/[1+(√3b/a)]=-b/a

> b/a-√3=-b/a-√3(b/a)^2

> √3(b/a)^2+2(b/a)-√3=0

> b/a=√3/3

> a=√3b………2）

Simultaneous (1)(2) yields: a 2 = 3, b 2 = 1

So, c 2 = a 2 + b 2 = 3 + 1 = 4

So, c = 2

Then, the eccentricity e=c a=2 3, and the hyperbolic equation is: x 2 3-y 2 = 1

Satisfied! o(∩_o~

It looks familiar

Solution 1: Since the hyperbolic equation is x 2-ay 2=1, the hyperbola focuses on the branch base on the x-axis, and the point p(1,0) must be the right vertex of the hyperbola, and the points q and r are all on the right branch of the hyperbola and distributed in the first place.

1. Four quadrants.

Let the q coordinate be (x,y), then the r coordinate is (x,-y) by pqr is a regular triangle.

x-1=（√3）|y|

y^2=(x-1)^2/3

Substituting it into a hyperbolic equation gets.

3-a)x^2+2ax-a-3=0

x-1)[(3-a)x+(a+3)]=0

The two roots are x=1 (i.e., the abscissa of point p), and x=(a+3) (a-3) because q and r are both to the right of x=1.

then (a+3) (a-3) 1 and a 0

A 3

That is, the value range of a is (3,+

Solution 2: Combination of numbers and shapes.

Since the hyperbolic equation is x 2-ay 2=1, the hyperbola focuses on the x-axis, the point p(1,0) must be the right vertex of the hyperbola, and the two asymptotic squares talk about x ( a)y=0(a 0).

From the hyperbolic symmetry, it can be seen that

If you want PQR to form a regular triangle, then the passage of the hyperbola.

1. The inclination angle of the asymptote of the three quadrants must be less than 30°

i.e. 0 k 3 3

So 0 1 a 3 3

The solution of a 3 is a value range of (3,+

Let a be the semi-solid axis of the hyperbola, as defined by the hyperbola.

pf2|-|pf1|=2a

If the projection of the center of the inscribed circle of the triangle pf1f2 on the transverse axis is a(x,0), this point is also the tangent point between the inscribed circle and the transverse axis. Let b and c be the tangent points of the inscribed circle and PF1 and PF2 respectively. Considering that the same point leads to the circle two tangents equally:

Then there are: pf2-pf1=(pc+cf2)-(pb+bf1)cf2-bf1=af2-f1a

c-x)-[x-(-c)]

2x=2ax=-a, so the abscissa of the center of the inscribed circle is -a, that is, above the intersection of the left branch of the hyperbola and the x-axis.

The known conditions of this question are incorrect, and it is useless to give a semi-solid axis a and a focal length of 2c.

From the inscription, it can be seen that when p is the intersection of the hyperbola and the x-axis, the point p is the intersection point of the tangent disturbance rock of p and the asymptotic line a, and the medium-sized comma point of b.

F1 is the left focus and F2 is the right focus.

Defined by hyperbola: | pf1|-|pf2| |=2a, and the title says that p is the right branch, so pf1 > pf2, i.e. |pf1|-|pf2|=2a( ) and |pf1|=2|pf2|, so substitution is obtain|pf2|=2a>c-a (distance from origin to f2), solution: 3a

6: Comprehensive Problems of Hyperbola - High School Mathematics First Round Review Lesson Plan: Chapter 8 Conic Curves. doc

..3) At that time, the equation is, which means that the center is at the origin, the focus is on the x-axis, and the vertex is the ellipse; (4) At that time, the equation is x2 + y2 = 1, which represents the unit circle; (5) At that time, the equation is, indicating that the center is at the origin, the focus is on the y-axis, and the vertex is .Conic curves, hyperbolas.

3) At that time, the equation is, which means that the center is at the origin, the focus is on the x-axis, and the vertex is the ellipse; (4) At that time, the equation is x2 + y2 = 1, which represents the unit circle; (5) At that time, the equation is, indicating that the center is at the origin, the focus is on the y-axis, and the vertex is .See.

Solution: easy to know, |pf2|=2a.Let the abscissa of p be x, then x a

From the second definition of the conic curve, we can see that (c a) [x-(a c)] = 2a===>x=3a²/c＞a.===>c/a＜3.

e＜3.∴e∈(1,3)

distance = |ab|/√a²+b²)=ab/c = 1/4）c+1（1/4）c+1]*c=ab≤(a²+b²)/2=c²/21/4）c+1≤c/2

Later, you solved it, because you don't know how to write (1 4) C+1.