JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 6)
The area (in sq. units) of an equilateral triangle
inscribed in the parabola y2 = 8x, with one of
its vertices on the vertex of this parabola, is :
$$256\sqrt 3 $$
$$64\sqrt 3 $$
$$128\sqrt 3 $$
$$192\sqrt 3 $$
Explanation
Let A = (2t2
, 4t)
B = (2t2 , -4t) (by symmetry as equilateral triangle)
_2nd_September_Evening_Slot_en_6_1.png)
tan 30o = $$\frac{4t}{2t^{2}} $$ = $$\frac{2}{t} $$
$$ \Rightarrow $$ t = 2$$\sqrt{3} $$
AB = 8t = 16$$\sqrt{3} $$
Area of equilateral = $$\frac{1}{2} \times 16\sqrt{3} \times 24$$
= $$192\sqrt 3 $$
B = (2t2 , -4t) (by symmetry as equilateral triangle)
tan 30o = $$\frac{4t}{2t^{2}} $$ = $$\frac{2}{t} $$
$$ \Rightarrow $$ t = 2$$\sqrt{3} $$
AB = 8t = 16$$\sqrt{3} $$
Area of equilateral = $$\frac{1}{2} \times 16\sqrt{3} \times 24$$
= $$192\sqrt 3 $$
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