Studdy AI Logo
Math

Question原点Oと点P(-3/2, 0)があり、曲線y=36x2y=\frac{\sqrt{3}}{6} x^{2}上の点A, B, Cを求める問題。 (1) A, B, Cの座標 (2) 円の中心座標 (3) Bの接線の式 (4) Bのみが両方にあることを示せ。

Studdy Solution

STEP 1

Assumptions1. The origin is denoted by O. The coordinates of are (−3,0)\left(-\frac{3}{},0\right)3. The points A, B, C are on the graph of y=36xy=\frac{\sqrt{3}}{6} x^{}
4. ∠OPA=∠OPB=30∘\angle \mathrm{OPA}=\angle \mathrm{OPB}=30^{\circ} and ∠ABC=60∘\angle \mathrm{ABC}=60^{\circ}

STEP 2

First, we need to find the coordinates of points A and B. Since ∠OPA=∠OPB=30∘\angle \mathrm{OPA}=\angle \mathrm{OPB}=30^{\circ}, the coordinates of A and B can be found by rotating the point around the origin by 30∘30^{\circ} and −30∘-30^{\circ} respectively.

STEP 3

The rotation of a point (x,y)(x, y) around the origin by an angle θ\theta is given by the following transformation(x′,y′)=(xcos⁡θ−ysin⁡θ,xsin⁡θ+ycos⁡θ)(x', y') = (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)

STEP 4

Substitute the coordinates of and the angle into the transformation to find the coordinates of A.
A=(−32cos⁡30∘−0sin⁡30∘,−32sin⁡30∘+0cos⁡30∘)A = \left(-\frac{3}{2} \cos30^{\circ} -0 \sin30^{\circ}, -\frac{3}{2} \sin30^{\circ} +0 \cos30^{\circ}\right)

STEP 5

Calculate the coordinates of A.
A=(−32cos⁡30∘,−32sin⁡30∘)A = \left(-\frac{3}{2} \cos30^{\circ}, -\frac{3}{2} \sin30^{\circ}\right)

STEP 6

Similarly, substitute the coordinates of and the angle into the transformation to find the coordinates of B.
B=(−32cos⁡−30∘−0sin⁡−30∘,−32sin⁡−30∘+0cos⁡−30∘)B = \left(-\frac{3}{2} \cos -30^{\circ} -0 \sin -30^{\circ}, -\frac{3}{2} \sin -30^{\circ} +0 \cos -30^{\circ}\right)

STEP 7

Calculate the coordinates of B.
B=(−32cos⁡−30∘,−32sin⁡−30∘)B = \left(-\frac{3}{2} \cos -30^{\circ}, -\frac{3}{2} \sin -30^{\circ}\right)

STEP 8

Since ∠ABC=60∘\angle \mathrm{ABC}=60^{\circ}, the coordinates of C can be found by rotating the point B around the point A by 60∘60^{\circ}. The rotation of a point (x,y)(x, y) around another point (a,b)(a, b) by an angle θ\theta is given by the following transformation(x′,y′)=((x−a)cos⁡θ−(y−b)sin⁡θ+a,(x−a)sin⁡θ+(y−b)cos⁡θ+b)(x', y') = ((x-a) \cos \theta - (y-b) \sin \theta + a, (x-a) \sin \theta + (y-b) \cos \theta + b)

STEP 9

Substitute the coordinates of A and B and the angle into the transformation to find the coordinates of C.
C=((Bx−Ax)cos⁡60∘−(By−Ay)sin⁡60∘+Ax,(Bx−Ax)sin⁡60∘+(By−Ay)cos⁡60∘+Ay)C = ((B_x-A_x) \cos60^{\circ} - (B_y-A_y) \sin60^{\circ} + A_x, (B_x-A_x) \sin60^{\circ} + (B_y-A_y) \cos60^{\circ} + A_y)

STEP 10

Calculate the coordinates of C.
C=((Bx−Ax)cos⁡60∘−(By−Ay)sin⁡60∘+Ax,(Bx−Ax)sin⁡60∘+(By−Ay)cos⁡60∘+Ay)C = ((B_x-A_x) \cos60^{\circ} - (B_y-A_y) \sin60^{\circ} + A_x, (B_x-A_x) \sin60^{\circ} + (B_y-A_y) \cos60^{\circ} + A_y)

STEP 11

The center of the circle passing through the points A, B, C can be found by solving the system of equations formed by the perpendicular bisectors of the segments AB and BC.

STEP 12

The equation of the perpendicular bisector of a segment with endpoints (x,y)(x, y) and (x2,y2)(x2, y2) is given byy−y+y22=−x2−xy2−y(x−x+x22)y - \frac{y + y2}{2} = -\frac{x2 - x}{y2 - y}(x - \frac{x + x2}{2})

STEP 13

Substitute the coordinates of A and B into the equation to find the equation of the perpendicular bisector of AB.
y−Ay+By2=−Bx−AxBy−Ay(x−Ax+Bx2)y - \frac{A_y + B_y}{2} = -\frac{B_x - A_x}{B_y - A_y}(x - \frac{A_x + B_x}{2})

STEP 14

Similarly, substitute the coordinates of B and C into the equation to find the equation of the perpendicular bisector of BC.
y−By+Cy2=−Cx−BxCy−By(x−Bx+Cx2)y - \frac{B_y + C_y}{2} = -\frac{C_x - B_x}{C_y - B_y}(x - \frac{B_x + C_x}{2})

STEP 15

olve the system of equations formed by the two perpendicular bisectors to find the coordinates of the center of the circle.

STEP 16

The equation of the tangent line to the circle at a point (x,y)(x, y) with center (h,k)(h, k) is given byy−y=x−x(x−h)2+(y−k)2y - y = \frac{x - x}{\sqrt{(x - h)^2 + (y - k)^2}}

STEP 17

Substitute the coordinates of B and the center of the circle into the equation to find the equation of the tangent line at B.
y−By=x−Bx(Bx−h)2+(By−k)2y - B_y = \frac{x - B_x}{\sqrt{(B_x - h)^2 + (B_y - k)^2}}

STEP 18

The point B is the only point on both the graph of y=36x2y=\frac{\sqrt{3}}{6} x^{2} and the tangent line. This can be shown by substituting the equation of the tangent line into the equation of the graph and showing that the only solution is the coordinates of B.

STEP 19

Substitute the equation of the tangent line into the equation of the graph.
\frac{\sqrt{3}}{6} x^{} = B_y + \frac{x - B_x}{\sqrt{(B_x - h)^ + (B_y - k)^}}

STEP 20

olve the equation for x.
x = \sqrt{\frac{6 (B_y + \frac{x - B_x}{\sqrt{(B_x - h)^ + (B_y - k)^}})}{\sqrt{3}}}

STEP 21

The only solution is x=Bxx = B_x, which shows that B is the only point on both the graph and the tangent line.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord