Find all values ​​of parameter a for each. Solving a problem with parameters

Find all values ​​of a, for each of which the system of equations

has exactly two solutions.

Solution.

Let's write the 1st equation of the system in the form: x 2 + 5x + y 2 -y -52 = |x-5y +5|. (*)

1) Since right side equality is non-negative, then the left side of the equality must be so, namely: x 2 + 5x + y 2 -y-52 ≥ 0. Let us select from algebraic sums(x 2 + 5x) and (y 2 - y) perfect squares binomials.

x 2 + 2 ∙ X ∙ 2.5 + 2.5 2 -2.5 2 + y 2 -2∙y∙0.5 + 0.5 2 -0.5 2 -52 ≥ 0;

(x 2 + 2 ∙ X ∙ 2.5 + 2.5 2) + (y 2 -2 ∙ y ∙ 0,5 + 0,5 2) ≥ 52 + 2,5 2 + 0,5 2 ;

(x + 2.5) 2 + (y-0.5) 2 ≥ 52 + 6.25 + 0.25;

(x + 2.5) 2 + (y-0.5) 2 ≥ 58.5. ODZ: solutions of the system are found among the set of points lying outside the circle with the center at the point Q(-2.5, 0.5) and radius

2) Let us expand the modular brackets in equation (*), assuming that the expression under the modulus sign is non-negative, i.e. x-5y +5 ≥ 0 or 5y ≤ x + 5, hence y ≤ 0.2x+1. Then equality (*) will be written as:

x 2 + 5x + y 2 -y-52 = x-5y +5. Let's move everything to left side and let's simplify it.

x 2 + 5x + y 2 -y-52-x + 5y-5 = 0;

x 2 + 4x + y 2 + 4y-57 = 0. Let us select the complete squares of binomials from the algebraic sums (x 2 + 4x) and (y 2 + 4y).

x 2 + 4x + 4-4 + y 2 + 4y +4-4-57 = 0;

(x 2 + 4x + 4) + (y 2 + 4y +4) = 57 + 4 + 4;

(x + 2) 2 + (y + 2) 2 = 65. This is the equation of a circle with center at point O 1 (-2; -2) and radius

We will consider only those points of this circle that lie below the straight line x-5y +5 = 0, since we obtained the equation of this circle under the condition that x-5y +5 ≥ 0, i.e. at y ≤ 0.2x+1. Note that all points of this circle lying below the straight line x-5y +5 = 0 are located outside the circle with the center at the pointQ(-2.5; 0.5), therefore satisfy the ODZ.

3) Now let’s open the modular brackets in equation (*), assuming that the expression under the modulus sign is negative, i.e. x-5y +5< 0 или 5у >x + 5, hence y>0.2x+1. Then equality (*) will be written as:

x 2 + 5x + y 2 -y-52 = -x + 5y +5. Let's move everything to the left side and simplify it.

x 2 + 5x + y 2 -y-52 + x-5y + 5 = 0;

x 2 + 6x + y 2 -6y-47 = 0. Let us select the complete squares of binomials from the algebraic sums (x 2 + 6x) and (y 2 -6y).

x 2 + 6x + 9-9 + y 2 -6y + 9-9-47 = 0;

(x 2 + 6x + 9) + (y 2 -6y +9) = 47 + 9 + 9;

(x + 3) 2 + (y-3) 2 = 65. This is the equation of a circle with center at point O 2 (-3; 3) and radius

We will consider only those points of this circle that lie above the straight line x-5y +5 = 0, since we obtained the equation of this circle under the condition x-5y +5< 0, т.е. при условии у >0.2x+1. Note that all points of this circle lying above the straight line x-5y +5 = 0 are located outside the circle with the center at the pointQ(-2.5; 0.5), therefore they satisfy the ODZ.

4) Find the intersection points of the circles with centers at points O 1 and O 2. These are also the points of intersection of any of these circles with the straight line x-5y +5 = 0. To be specific, let’s take the equation of the first of the circles and solve the system:

From the 2nd equation, we express x through y and substitute it into the 1st equation.

Let us simplify and solve the 2nd equation of the resulting system.

(5y-3) 2 + (y + 2) 2 = 65;

25y 2 -30y + 9 + y 2 +4y + 4-65 = 0;

26y 2 -26y-52 = 0;

y 2 -y-2 = 0. According to Vieta’s theorem, y 1 + y 2 = 1, y 1 ∙ y 2 = -2. Hence y 1 = -1, y 2 = 2.

Then x 1 = 5 ∙ y 1 -5 = 5 ∙ (-1)-5 = -10; x 2 = 5 ∙ y 2 -5 = 5 ∙ 2-5 = 2.

The intersection points of the circles with centers O 1 and O 2 lie on the straight line x-5y +5 = 0, and these are points T(-10; -1) and A(5; 2).

5) Let’s figure out what the straight line y-2 = a(x-5) is. Let's write this equation in the form y = a(x-5) + 2 and remember how it turns out graph of a functiony = f(x-m) + nfrom the graph of a functiony = f(x). It is obtained by transferring the graph of the functiony = f(x) onmsingle segments along the Ox axis and onnsingle segments along the Oy axis. Therefore, the graph of the function y = a(x-5) + 2 can be obtained from the graph of the function y = ax by moving 5 units to the right and 2 units up. In other words, the straight line will pass through point A(5; 2) and must have the following slope A, to intersect our circles with centers at points O 1 and O 2 in exactly two points. This will happen only in those cases when the straight line, passing through point A, common to both circles, will then intersect only one of them. The limit positions of our straight line (with the parameter A) will be tangents to the circles at point A. We will need not the equations of the tangents themselves, but their slopes. How will we get them?

6) Radius O 1 A drawn to the point of contact will be perpendicular to the tangent. Angle coefficientsk 1 Andk 2 two mutually perpendicular linesy = k 1 x+ b 1 Andy = k 2 x+ b 2 obey the law:k 1 ∙ k 2 = -1. Let's compose the equations of straight line O 1 A and straight line O 2 A, determine the angular coefficient of each straight line, and then find the angular coefficients of the tangents, which are the limiting positions of the straight line y = a(x-5) + 2. The gap between the found values ​​of the parameter A and will be the answer to the problem.

We use the formula for the equation of a straight line passing through two given points (x 1; y 1) and (x 2; y 2). This formula looks like:

Let's create an equation for a straight line passing through points O 1 (-2; -2) and A (5; 2). We have x 1 = -2, y 1 = -2, x 2 = 5, y 2 = 2. Substitute these values ​​into the formula:

So, the equation of the tangent at point A to a circle with center at point O 1 has the form.

1. Task.
At what parameter values a equation ( a - 1)x 2 + 2x + a- Does 1 = 0 have exactly one root?

1. Solution.
At a= 1 the equation is 2 x= 0 and obviously has a single root x= 0. If a No. 1 then given equation is square and has a single root for those parameter values ​​for which the discriminant quadratic trinomial equal to zero. Equating the discriminant to zero, we obtain an equation for the parameter a 4a 2 - 8a= 0, whence a= 0 or a = 2.

1. Answer: the equation has a single root at a O (0; 1; 2).

2. Task.
Find all parameter values a, for which the equation has two different roots x 2 +4ax+8a+3 = 0.
2. Solution.
Equation x 2 +4ax+8a+3 = 0 has two distinct roots if and only if D = 16a 2 -4(8a+3) > 0. We get (after reduction by a common factor of 4) 4 a 2 -8a-3 > 0, whence

2. Answer:

3. Task.
It is known that
f 2 (x) = 6x-x 2 -6.
a) Graph the function f 1 (x) at a = 1.
b) At what value a function graphs f 1 (x) And f 2 (x) have a single common point?

3. Solution.
3.a. Let's transform f 1 (x) as follows
The graph of this function at a= 1 is shown in the figure on the right.
3.b. Let us immediately note that the graphs of functions y = kx+b And y = ax 2 +bx+c (a No. 0) intersect at a single point if and only if the quadratic equation kx+b = ax 2 +bx+c has a single root. Using View f 1 of 3.a, let us equate the discriminant of the equation a = 6x-x 2 -6 to zero. From equation 36-24-4 a= 0 we get a= 3. Do the same with equation 2 x-a = 6x-x 2 -6 we will find a= 2. It is easy to verify that these parameter values ​​satisfy the conditions of the problem. Answer: a= 2 or a = 3.

4. Task.
Find all values a, for which the set of solutions to the inequality x 2 -2ax-3a i 0 contains the segment .

4. Solution.
First coordinate of the parabola vertex f(x) = x 2 -2ax-3a equal to x 0 = a. From properties quadratic function condition f(x) i 0 on the segment is equivalent to a set of three systems
has exactly two solutions?

5. Solution.
Let us rewrite this equation in the form x 2 + (2a-2)x - 3a+7 = 0. This is a quadratic equation, it has exactly two solutions if its discriminant is strictly greater than zero. Calculating the discriminant, we find that the condition for the presence of exactly two roots is the fulfillment of the inequality a 2 +a-6 > 0. Solving the inequality, we find a < -3 или a> 2. The first of the inequalities is obviously solutions in natural numbers does not have, and the smallest natural solution to the second is the number 3.

5. Answer: 3.

6. Problem (10 keys)
Find all values a, for which the graph of the function or, after obvious transformations, a-2 = | 2-a| . The last equation is equivalent to the inequality a i 2.

6. Answer: a ABOUT )