Points A and B are selected on the graph of x2+2 y=0 so that the triangle A B O is equilateral. The length of the side of the triangle is ('O' is the origin)

Question

Points A and B are selected on the graph of x2+2 y=0 so that the triangle A B O 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

Tardigrade Admin · Accepted Answer

y=-(x2/2) tan 30°=(2 t / t 2) ⇒ (1/√3)=(2/ t ) ⇒ t =2 √3 l( OP )=4 √3

Q. Points $A$ and $B$ are selected on the graph of $x^{2} + 2 y = 0$ so that the triangle $A BO$ is equilateral. The length of the side of the triangle is ('O' is the origin)

$43$

$\frac{4}{3}$

$\frac{4}{3}$

$23$

$y = - \frac{x ^{2}}{2}$
$tan 3 0^{\circ} = \frac{2 t}{t ^{2}} \Rightarrow \frac{1}{3} = \frac{2}{t} \Rightarrow t = 23$
$l (OP) = 43$

Q. 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)

43​

34​

3​4​

23​

y=−2x2​ tan30∘=t22t​⇒3​1​=t2​⇒t=23​ l(OP)=43​

Q. Points $A$ and $B$ are selected on the graph of $x^{2} + 2 y = 0$ so that the triangle $A BO$ is equilateral. The length of the side of the triangle is ('O' is the origin)

$43$

$\frac{4}{3}$

$\frac{4}{3}$

$23$

$y = - \frac{x ^{2}}{2}$
$tan 3 0^{\circ} = \frac{2 t}{t ^{2}} \Rightarrow \frac{1}{3} = \frac{2}{t} \Rightarrow t = 23$
$l (OP) = 43$