In the xy-plane, the line y = mx, where m is a positive constant, is tangent to the parabo… — SAT

In the $xy$ -plane, the line $y = mx$ , where $m$ is a positive constant, is tangent to the parabola $y = x^2 + 9$ . What is the value of $m$ ?

The value of

m

is 6.

Equating the equations of the line and the parabola yields the quadratic equation

x^2 - mx + 9 = 0

. For the line to be tangent to the parabola, this system must have exactly one real solution, meaning the discriminant

b^2 - 4ac

must equal

0

. Substituting

a = 1

b = -m

, and

c = 9

into the discriminant formula gives

(-m)^2 - 4(1)(9) = 0

, which simplifies to

m^2 - 36 = 0

. Solving for the positive constant

m

gives

6

Set the equation of the line equal to the equation of the parabola to find their intersection points.

mx = x^2 + 9

The intersection points of the system correspond to the values of

x

where both equations have the same

y

-value.

Rewrite the equation in standard quadratic form

ax^2 + bx + c = 0

by subtracting

mx

from both sides.

x^2 - mx + 9 = 0

This allows the identification of the coefficients

a

b

, and

c

to compute the discriminant.

Identify the coefficients and set the discriminant

b^2 - 4ac

equal to 0.

(-m)^2 - 4(1)(9) = 0

A line is tangent to a parabola if and only if the system has exactly one real solution, which corresponds to a discriminant of zero.

Solve the equation for the positive constant

m

m^2 - 36 = 0 \implies m^2 = 36 \implies m = 6

Taking the square root of both sides gives

m = \pm 6

. Since

m

is specified as a positive constant, we choose

m = 6

Solving nonlinear systems of equations where a line is tangent to a parabola by setting the discriminant of the resulting quadratic equation to zero.

Question