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An equilateral triangle is inscribed in the parabola `y^2=4ax` whose vertex is at of the parabola. Find the length of its side.

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To find the length of the side of the equilateral triangle inscribed in the parabola \(y^2 = 4ax\) with one vertex at the vertex of the parabola, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Vertex of the Parabola**: The vertex of the parabola \(y^2 = 4ax\) is at the point \((0, 0)\). **Hint**: Recall that the vertex of the parabola in the form \(y^2 = 4ax\) is always at the origin. ...
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Knowledge Check

  • An equilateral trinagle is inscribed in parabola y^2=8x whose one vertex coincides with vertex of parabola.Find area of triangle.

    A
    `(196sqrt3)`
    B
    `(194sqrt3)`
    C
    `(192sqrt3)`
    D
    `(190sqrt3)`

  • If OAB is an equilateral triangle inscribed in the parabola y^(2) = 4ax with O as the vertex, then the length of the side of triangle OAB is

    A
    `8asqrt(3)`
    B
    `4asqrt(3)`
    C
    `2asqrt(3)`
    D
    `asqrt(3)`

  • If OAB is an equilateral triangle inscribed in the parabola y^(2)=4ax with O as the vertex, then the length of the side of the DeltaOAB is

    A
    `8a sqrt(3)`
    B
    `4a sqrt(3)`
    C
    `2a sqrt(3)`
    D
    `a sqrt(3)`

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