Question 8 An equilateral triangle is inscribed in the parabola y^2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Class X1 - Maths -Conic Sections Page 264
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first of all we have to draw parabola in the positive direction of X - axis and inside that draw an equilateral triangle OAB ( Let )
Let OB = OA = AB = l
we know, one thing parabola is always symmetry about own axis .
so, angle BOA of triangle equally divided by axis of parabola e.g X - axis.
so, ∠BOP =∠POA = 30°
in ∆BOP,
use sin∅ = perpendicular/hypotenuse
sin30° = PB/OB
1/2 = PB/OB
1/2 = PB/l
PB = l/2
again, use cos∅ = base/hypotenuse
cos∅ = OP/OB
cos30° = OP/OB
√3/2 = OP/l
OP = l√3/2
hence, co-ordinate of B = (OP, BP )
B = ( l√3/2, l/2 ) this point lies on parabola .
so, it will satisfy y² = 4ax .
(l/2)² = 4a × l√3/2 = 2√3al
l²/4 = 2√3al
l = 8√3a
hence, Length of the side of equilateral ∆ = 8√3a
Let OB = OA = AB = l
we know, one thing parabola is always symmetry about own axis .
so, angle BOA of triangle equally divided by axis of parabola e.g X - axis.
so, ∠BOP =∠POA = 30°
in ∆BOP,
use sin∅ = perpendicular/hypotenuse
sin30° = PB/OB
1/2 = PB/OB
1/2 = PB/l
PB = l/2
again, use cos∅ = base/hypotenuse
cos∅ = OP/OB
cos30° = OP/OB
√3/2 = OP/l
OP = l√3/2
hence, co-ordinate of B = (OP, BP )
B = ( l√3/2, l/2 ) this point lies on parabola .
so, it will satisfy y² = 4ax .
(l/2)² = 4a × l√3/2 = 2√3al
l²/4 = 2√3al
l = 8√3a
hence, Length of the side of equilateral ∆ = 8√3a
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