Solve this: Q 66 Points A and B are selected on the graph of x2 + 2y = 0 - Maths - Linear Inequalities

Question

Solve this:
Q 66 Points A and B are selected on the graph of x2 + 2y = 0 so
that the triangle ABO is equilateral The length of the side of the
triangle is ('O' is the origin)
(A) 4 3
(B) 4 3
(C) 4 3
(D) 2 3

Ghanshyam Dhakar · Accepted Answer

Dear student,
graph x2+2y = 0, y=-x22let point Ax1,-x122,   Bx2,-x222 and O0,0for equilateral Triangle OA=OBx12+x144=x22+x244x12+x144=x22+x244x12-x22+x144-x244=0x12-x22+x12-x22x124+x224=0x12-x221+x124+x224=0then x12-x22=0x12=x22x1=-x2 and x1=+x2 AB=OAx1-x22+x122-x222 = x12+x144x1-x22+x122-x222 = x12+x144use  x1=+x2 0=  x12+x144   not possibleso use  x1=-x2 2x12+x122-x122 = x12+x1444x12 = x12+x1443x12=x1443=x124x12 =12x1 =12 , -12use x1 =12point  Ax1,-x122=12,-6then OA = 12-02+-6-02=48=43  answer
Regards

Solve this: Q.66. Points A and B are selected on the graph of x2 + 2y = 0 so that the triangle ABO is equilateral. The length of the side of the triangle is ('O' is the origin) (A) 4 3 (B) 4 3 (C) 4 3 (D) 2 3

Solve this:
Q.66. Points A and B are selected on the graph of x2 + 2y = 0 so that the triangle ABO is equilateral. The length of the side of the triangle is ('O' is the origin)
(A) $4 \sqrt{3}$
(B) $\frac{4}{3}$
(C) $\frac{4}{\sqrt{3}}$
(D) $2 \sqrt{3}$