Binomial Theorem

Question
CBSEENMA11015132

An equilateral triangle is inscribed in the parabola straight y squared equals 4 ax., where one vertex is at the vertex of the parabola. Find (a)  the length of the side of the triangle, (b) area of triangle ABC.

Solution

Let OAB be the triangle with side p.
In straight capital delta space OLA comma space OA space equals space straight p
Now,               OL over straight p space equals space cos space 30 degree space equals space fraction numerator square root of 3 over denominator 2 end fraction space rightwards double arrow space OL space equals space fraction numerator square root of 3 over denominator 2 end fraction straight p
and   AL over straight p space equals space sin space 30 degree space equals space 1 half space rightwards double arrow space AL space equals space 1 half straight p
∴ Point A is open parentheses fraction numerator square root of 3 over denominator 2 end fraction straight p comma space 1 half straight p close parentheses
Since it lies on straight y squared equals 4 ax
∴    open parentheses 1 half straight p close parentheses squared space equals space 4 straight a open parentheses fraction numerator square root of 3 over denominator 2 end fraction straight p close parentheses space rightwards double arrow space 1 fourth straight p squared space equals space 2 straight a square root of 3 straight p space rightwards double arrow space straight p equals 8 square root of 3 straight a
Hence, (i) each side of space space space space space space ΔABC space equals straight p equals space 8 square root of 3 straight a,    (ii) area of ΔABC space equals space fraction numerator square root of 3 straight p squared over denominator 4 end fraction equals fraction numerator square root of 3 over denominator 4 end fraction left parenthesis 64 cross times 3 straight a squared right parenthesis space equals space 48 square root of 3 straight a squared
                                            

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