In the standard (x, y) coordinate plane, a circle is defined by the equation x^2 + y^2 - 1… — ACT

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 - 12y + 27 = 0$ . A parabola that opens downward has its vertex at $(0, k)$ and is defined by the equation $y = -x^2 + k$ . If the system of equations consisting of this circle and parabola has exactly three distinct real solution points, what is the value of $k$ ?

Cevap: 9

Cevap

The value of

k

is 9.

The correct value of

k

is 9 because when

k=9

, the system of equations reduces to a quadratic in

y

with roots

y=9

and

y=4

. Both roots satisfy the real-number constraint

y \leq 9

for the parabola

x^2 = 9-y

, producing three distinct real solutions:

(0, 9)

(\sqrt{5}, 4)

, and

(-\sqrt{5}, 4)

Adım Adım Çözüm

Complete the square for the circle's equation.

x^2 + (y-6)^2 = 9

To identify the circle's center at

(0, 6)

and radius

R=3

for geometric interpretation.

Express

x^2

in terms of

y

using the parabola's equation.

x^2 = k - y

To substitute into the circle's equation and eliminate the

x

variable.

Substitute

x^2

into the circle's equation and simplify.

y^2 - 13y + (k+27) = 0

To create a quadratic equation in

y

representing the y-coordinates of the intersection points.

Set

y = k

in the quadratic equation.

k^2 - 12k + 27 = 0

, which factors as

(k-3)(k-9) = 0

An intersection must lie on the y-axis (

x=0

, which means

y=k

) to yield an odd number of intersection points.

Verify which candidate value of

k

yields exactly three real solutions.

For

k=3

, the solutions are restricted because

y=10

gives no real

x

value, resulting in only 1 solution. For

k=9

, the roots

y=9

and

y=4

both yield real

x

values, resulting in exactly 3 solutions:

(0, 9)

(\sqrt{5}, 4)

, and

(-\sqrt{5}, 4)

The algebraic condition for real

x

coordinates is

x^2 = k - y \geq 0

, so we must verify that the roots

y

satisfy

y \leq k

Anahtar Kavram

Solving systems of non-linear equations algebraically and analyzing the number of real intersection points under coordinate constraints.

Alternatif Yöntem

Geometrically, a parabola opening downward with its vertex on the y-axis will intersect a circle centered on the y-axis in exactly three points if and only if its vertex is at the top of the circle and its curvature is less than that of the circle at that point. Completing the square for the circle

x^2 + y^2 - 12y + 27 = 0

gives

x^2 + (y-6)^2 = 9

, which shows the top point of the circle is

(0, 9)

. Thus, the vertex of the downward-opening parabola must be at

(0, 9)

, meaning

k = 9

. We then algebraically verify that this curvature indeed allows two other real intersections.

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Cevap

Adım Adım Çözüm

Anahtar Kavram

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Alternatif Yöntem

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