Question

Ratio of the area cut off a parabola by any double ordinate is that of the corresponding rectangle contained by that double ordinate and its distance from the vertex is -

• A. ½
• B. 1/3
• C. 2/3
• D. 1

Answer 1

Answer: (C)

2/3

Answer 2

Let y² = 4ax be a parabola and let x = b be a double ordinate. Then, A₁ = Area enclosed by the parabola y² = 4ax and the double ordinate x = b.

= 2 $\int _ { 0 } ^ { b } y$ dx = 2 $\int _ { 0 } ^ { b } \sqrt { 4 a x }$ dx = 4 $\sqrt { \mathrm { a } }$ $\int _ { 0 } ^ { b } \sqrt { x }$ dx

= 4 $\sqrt { \mathrm { a } }$ $\left[ \frac { 2 } { 3 } x ^ { 3 / 2 } \right] _ { 0 } ^ { b }$

= 4 $\sqrt { \mathrm { a } }$ × $\frac { 2 } { 3 } b ^ { 3 / 2 }$= $\frac { 8 } { 3 } a ^ { 1 / 2 } b ^ { 3 / 2 }$ And, A₂

= Area of the rectangle ABCD = AB × AD =$2 \sqrt { 4 a b }$× b = 4 a^1/2 b^3/2

\ A₁ : A₂ = 8/3 a^1/2 b^3/2 : 4 a^1/2 b^3/2 = 2/3 : 1 = 2 : 3.