Question 2 An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?
Class X1 - Maths -Conic Sections Page 264
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see attachment, axis is vertical , so let arc of parabola is in the form of x² = 4ay
here, OB = 10 m
AC = 5 m so, AB = BC = 5/2 = 2.5 m
so, the co-ordinate of A = ( 2.5 , 10)
point A ( 2.5, 10) will satisfy equation of parabola .
so,
( 2.5)² = 4 × a × 10
6.25 = 40a
a = 6.25/40 = 5/32
now, equation of parabola is x² = 4 × 5/32y
x² = 5y/8 __________(1)
Let OR = 2 m and PQ = k then, RP = k/2
hence, point P( k/2, 2) .
point P will also satisfy equation of parabola .
put P ( k/2, 2) in equation (1)
(k/2)² = 5 × 2/8
k²/4 = 5/4
k = ±√5 but k is distance so, K won't be negative.
k = √5 .
hence, √5 m wide it is 2m from the vertex of the parabola.
here, OB = 10 m
AC = 5 m so, AB = BC = 5/2 = 2.5 m
so, the co-ordinate of A = ( 2.5 , 10)
point A ( 2.5, 10) will satisfy equation of parabola .
so,
( 2.5)² = 4 × a × 10
6.25 = 40a
a = 6.25/40 = 5/32
now, equation of parabola is x² = 4 × 5/32y
x² = 5y/8 __________(1)
Let OR = 2 m and PQ = k then, RP = k/2
hence, point P( k/2, 2) .
point P will also satisfy equation of parabola .
put P ( k/2, 2) in equation (1)
(k/2)² = 5 × 2/8
k²/4 = 5/4
k = ±√5 but k is distance so, K won't be negative.
k = √5 .
hence, √5 m wide it is 2m from the vertex of the parabola.
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