An equilateral triangle is inscribed in the p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $a$ be the length of the side of the equilateral triangle.Since the triangle is symmetric about the $x$-axis,one vertex is at the origin $(0,

Vedclass · Accepted Answer

$16 \sqrt{3} 	ext{ units}$

Vedclass · Answer

(B) Let $a$ be the length of the side of the equilateral triangle.Since the triangle is symmetric about the $x$-axis,one vertex is at the origin $(0, 0)$.The other two vertices are at $(x, y)$ and $(x, -y)$.For an equilateral triangle with side $a$,the height is $\frac{\sqrt{3}}{2}a$ and the vertical distance from the $x$-axis to the vertices is $\frac{a}{2}$.Thus,the coordinates of the vertex on the parabola are $\left(\frac{\sqrt{3}}{2}a, \frac{a}{2}ight)$.Since this point lies on the parabola $y^2 = 8x$,we substitute the coordinates:$\left(\frac{a}{2}ight)^2 = 8 \left(\frac{\sqrt{3}}{2}aight)$$\frac{a^2}{4} = 4\sqrt{3}a$Since $a 
eq 0$,we divide by $a$:$\frac{a}{4} = 4\sqrt{3}$$a = 16\sqrt{3} 	ext{ units}$.

An equilateral triangle is inscribed in the parabola $y^2 = 8x$,with one of its vertices at the vertex of the parabola. Then,the length of the side of that triangle is

Similar Questions

What is the length of the subnormal at any point on the parabola $y^2 = 4ax$?

The length of the latus rectum of the parabola $2\{(x - 1)^2 + (y - 3)^2\} = (x + y - 1)^2$ is:

If the locus of a point that divides a chord of slope $2$ of the parabola $y^2 = 4x$ internally in the ratio $1:2$ is a parabola,then its vertex is

If $x + y = K$ is a normal to the parabola $y^2 = 12x$,then what is the value of $K$?

The point of intersection of the latus rectum and the axis of the parabola ${y^2} + 4x + 2y - 8 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module