Question

A line has equation 2x + y=20 and a curve has equation y=a+

Answer 1

Answer:

we have two equations:

2*x + y = 20

y = a + 18/x - 3

We want to find the values of a such that the graphs of the equations do not intersect.

The first step is to isolate y in the first equation.

now we get:

y = 20 - 2*x

Then we have the system of equations:

y = 20 - 2*x

y = a + 18/x - 3

If the line intersects the curve at a point (x, y), then at that point both functions have the same value of y, then:

20 - 2*x = y = a + 18/x- 3

We want to find a value of such that this does not happen.

Let's solve that equation and see what we can find.

20 - 2*x =  a + 18/x - 3

20 + 3 = a + 18/x + 2*x

23 = a + 18/x + 2*x

(23 - a) = 18/x + 2*x

if we multiply all values by x, we get:

(23 - a)*x = 18 + 2*x^2

Then we have a quadratic equation, we can write this as:

2*x^2 - (23 - a)*x + 18 = 0

Now, remember that if the determinant of the quadratic function is smaller than zero, then the function does not have any real solution, then we need to find a such that the determinant of this quadratic equation is smaller than zero.

Remember that for a quadratic equation:

a*x^2 + b*x + c = 0 is:

b^2 - 4*a*c

In this case, the determinant is:

(-(23 - a))^2 - 4*2*18

(23 - a)^2 - 144

And this must be smaller than zero, then:

(23 - a)^2 - 144 < 0

23^2 - 2*23*a + a^2 - 144 < 0

529 - 46*a + a^2 - 144 < 0

a^2 - 46*a + 385 < 0

Notice that the function has a positive leading coefficient, then the arms of the function go upwards, then the region of the function that is smaller than zero is the values of a that are between both roots of a.

Then we need to solve:

a^2 - 46*a + 385 = 0

We can use the Bhaskara's equation to find the roots, the roots are given by:

$a = \frac{-(-46) +- \sqrt{(-46)^2 - 4*1*385} }{2*1} = \frac{46 +-24}{2}$

Then the two roots are:

a₁ = (46 + 24)/2 = 35

a₂ = (46 - 24)/2 = 11

Then for all the values of a suc that:

11 < a < 35

The line does not intersect the curve.