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Question: An equilateral triangle is inscribed in the parabola \({{y}^{2}}=4ax\), where one vertex is at the v...

An equilateral triangle is inscribed in the parabola y2=4ax{{y}^{2}}=4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
A. 83a8\sqrt{3}a
B. 43a4\sqrt{3}a
C. 23a2\sqrt{3}a
D. 3a\sqrt{3}a

Explanation

Solution

We first try to draw the parabola and the inscribed triangle. We assume the x value for the other two vertices and then find the coordinates of those vertices. We use the equal side length for the triangle to find the relation between hh and aa.

Complete step by step answer:
Let us draw the parabola y2=4ax{{y}^{2}}=4ax and the inscribed triangle where one vertex is at the vertex of the parabola. The general equation of parabola is (x−α)2=4a(y−β){{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right). For the general equation (α,β)\left( \alpha ,\beta \right) is the vertex. 4a is the length of the latus rectum. The coordinate of the focus is (α,β+a)\left( \alpha ,\beta +a \right). This gives the vertex as (0,0)\left( 0,0 \right). The length of the latus rectum is 4a4a. Therefore, one vertex of the triangle is A(0,0)A\left( 0,0 \right).

Let us take the other two vertices for x=hx=h. The vertices are on the parabola. So, putting the value of x=hx=h on y2=4ax{{y}^{2}}=4ax, we get y2=4ah{{y}^{2}}=4ah which gives y=±2ahy=\pm 2\sqrt{ah}.Therefore, the other two vertices are B(h,2ah)B\left( h,2\sqrt{ah} \right) and C(h,−2ah)C\left( h,-2\sqrt{ah} \right). Now this is an equilateral triangle which means all the sides are equal. So, AB=BCAB=BC.We know that the distance between two points (m,n)\left( m,n \right) and (o,p)\left( o,p \right) is (m−o)2+(n−p)2\sqrt{{{\left( m-o \right)}^{2}}+{{\left( n-p \right)}^{2}}}.

We find the length of the sides and get
AB=(h−0)2+(2ah−0)2=h2+4ahAB=\sqrt{{{\left( h-0 \right)}^{2}}+{{\left( 2\sqrt{ah}-0 \right)}^{2}}}=\sqrt{{{h}^{2}}+4ah}
⇒BC=(h−h)2+(2ah+2ah)2=16ah\Rightarrow BC=\sqrt{{{\left( h-h \right)}^{2}}+{{\left( 2\sqrt{ah}+2\sqrt{ah} \right)}^{2}}}=\sqrt{16ah}
The equation AB=BCAB=BC gives h2+4ah=16ah\sqrt{{{h}^{2}}+4ah}=\sqrt{16ah}. Simplifying we get

\Rightarrow {{h}^{2}}+4ah=16ah \\\ $$ Now we try to find the value of $h$ with respect to the value of $a$. As $h\ne 0$ we get $${{h}^{2}}+4ah=16ah \\\ \Rightarrow {{h}^{2}}=12ah \\\ \Rightarrow h=12a \\\ $$ Putting the value of $$h=12a$$ in $BC=\sqrt{16ah}$. We get $BC=\sqrt{16ah} \\\ \Rightarrow BC=\sqrt{16a\times 12a}$ $\Rightarrow BC=\sqrt{16a\times 12a} \\\ \therefore BC=8\sqrt{3}a$ **Hence, the correct option is A.** **Note:** We also can use the slope of the line AB. The angle for the equilateral triangle will be ${{60}^{\circ }}$. Therefore, the value of the slope $\dfrac{2\sqrt{ah}-0}{h-0}=\tan 60$. We simplify the equation and find the relation between $h$ and $a$ to use for the sides’ length.