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Q.
An equilateral triangle is inscribed in the parabola $y^2=4 a x$ whose one vertex is at the vertex of the parabola. Then, the length of the side of the triangle is
Conic Sections
Solution:
As shown in the figure $A P Q$ denotes the equilateral triangle with its equal side of length / (say).
Here, $A P=I \text { so } A R=I \cos 30^{\circ}$
Also, $ P R=I \sin 30^{\circ}=\frac{1}{2}$.
Thus, $\left(\frac{1 \sqrt{3}}{2}, \frac{1}{2}\right)$ are the coordinates of the point $P$ lying on the parabola $y^2=4 a x$.
Therefore, $\frac{I^2}{4}=4 a\left(\frac{I \sqrt{3}}{2}\right) \Rightarrow I=8 a \sqrt{3}$.
Thus, $8 a \sqrt{3}$ is the required length of the side of the equilateral triangle inscribed in the parabola $y^2=4 a x$.
