Lesson Geometric word problems that are solved using systems of non-linear equations

Geometric word problems that are solved using systems of non-linear equations

Problem 1

We look for intersection points of the circle y^2 = 10x-x^2 with a straight line
y = kx-2 and find "k" from the condition that two intersection points merge into one.


For it, we substitute y = kx-2 into the equation of the circle, and we get

    (kx-2)^2 = 10x - x^2

    k^2*x^2 - 4kx + 4 = 10x - x^2

    (k^2+1)x^2 - (4k+10) + 4 = 0.


Next, we calculate the discriminant of this quadratic equation

    d = b^2 - 4ac = [-(4k+10)]^2 - 16(k^2+1) = 16k^2 + 80k + 100 - 16k^2 - 16 = 80k + 84


and equate it to zero

    80k + 84 = 0.  


It gives us  

    80k = - 84,  k =  = .


ANSWER.  k =  = -1.05.

Problem 2

The standard form equation of the circle is

     +  = 3.           (1)


The equation of the line is

    y = -kx - 1.                      (2)



Substitute the equation (2) into equation (1), replacing y in the equation (1).

Doing this way, you will get then a single equation for only one unknown x
    
     +  = 3     (3)

or

     +  = 3.          (4)



The idea of the solution to the problem is to reduce equation (4) to standard form quadratic equation
and then to find the value of "k" from the condition that this equation has ONLY ONE real solution
(which is equivalent to the fact that the line(2) and the circle (1) have ONLY ONE common point).


So, we simplify equation (4) step by step

    x^2 - 4x + 4 + k^2x^2 + 4kx + 4 = 3

    x^2(1+k^2) - 4x(1-k) + 5 = 0.       (5)


Now we will find "k" from the condition that the discriminant of equation (5) is equal to zero.


The discriminant  d  is

    d = b^2 - 4ac =  -  =  - .


The condition d = 0  is

        = 

    4(1-k)^2       = 5(1+k^2)

    4 - 8k + 4k^2  = 5 + 5k^2

    k^2 + 8k + 1 = 0

     =  =  = .


So, there are 2 (two, TWO) values for k:   =   and   = .


ANSWER.  There are two solutions for "k" :   =   and   = .

Problem 3

To find the intersection points of the circle and the straight line,  substitute   =   into the equation   = .  You will get

     +  = ,


     +  = ,


     = ,


     = .


Next,  apply the quadratic formula to find the roots

     = ,


where    is the discriminant   =  =  = . 


Now everything is determined by the discriminant. 


    If  d > 0,  then there are two roots root. Correspondingly, there are two intersection points. 

    In opposite, if there are two intersection points, then there are two roots, hence, d > 0.


    If  d = 0,  then there is only one root. It means that the straight line tangents the circle. 

    And in opposite, if the straight line tangents the circle, then there is only one root; hence, d = 0.


    Finally, if  d < 0,  then there are no intersection points. 


Thus you need to solve this critical equation

     = 0.


It gives

    m = +/- . 


So, if   = +/- ,  then the straight line tangents the circle.


If  -sqrt(3)}}} < m < ,  then there are two intersection points. 


If  |m| > ,  then there are no intersection points. 

Problem 4

 = 36,   (1)
 = 4ky           (2)


The circle  =  has the center at (x,y) = (0,6) and has the radius of 6. 
So, the circle has the y-axis x=0 as a diameter and as a symmetry line, passes through the origin and touches the x-axis.

Parabola  =  also passes through the origin; has the y-axis x=0 as its symmetry line, and touches the x-axis.

After these geometric considerations (that are useful but are not absolutely necessary) let solve the problem algebraically.

Based on (2), substitute 4ky instead of  into the equation (1). You will get

 = 36.

So, in this way you excluded "x" from the system and got a single equation for "y". Let us simplify it:

 = 36,  or

 = 0.  (1)

Now, the problem requires this equation (1) to have only one non-negative solution.
     (One solution is evident/obvious. It is y = 0.)

It implies that (4k-12) MUST be non-positive:  4k-12 <= 0.
OTHERWISE y =  would be the other non-negative solution to (1).

So, the solution to the problem is this inequality 4k-12 <= 0,  or, equivalently,  k <= 3   (3 = )


Answer.  k <= 3.


See an illustration below for k = 3, 2, and 4.


    


    The circle  =  (red + green) 
    and three parabolas x^2 = 4ky for k=3 (blue), k=2, and k=4. 



    


    The circle  =  (green) 
    and three parabolas x^2 = 4ky for k=3 (blue), k=2, and k=4. 

Problem 5

An equation of the line passing through the point (7,5) is

    y-5 = m*(x-7),

where "m" is the slope coefficient, or

    y = mx - 7m +5.


    +----------------------------------------------------------------+
    |        The value of the slope "m" is unknown now,              |
    |    and the rest of the solution is to find the value of "m".   |
    +-----------------------------------------------------------==---+


Substitute this expression for y into the right side of the parabola formula

    x^2 = 6(mx - 7m +5) + 10.


You will get

    x^2 - 6mx + 42m - 40 = 0.    (1)


The line is tangent to the parabola if and only if the quadratic equation (1) has a unique real root.

It happens if and only if the discriminant of the equation is zero.


The discriminant is  d = b^2 - 4ac;  b= -6m;  a= 1;  c= 42m-40,  so

    d = 36m^2 - 4(42m-40) = 36m^2 - 168m + 160.


Therefore, to find "m", we need solve this quadratic equation

    36m^2 - 168m + 160 = 0.


Find its solutions, using the quadratic formula. They are

     =   and   = .


So, there are two tangent line through the given point to parabola.

Their equations are  y-5 =   and  y-5 = .


The equivalent forms of these equations are

    3y - 10x = -55  and  3y - 4x = -13.


The plots are shown in the Figure below.



    


    Parabola  y =  (red) and tangent lines 3y-4x = -13 (green)  and  3y-10x = -55 (blue)