Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_parameter

Class 12 Maths · Applications Of Integrals · Very Long Answer #49

For a>0, the region enclosed by y²=16ax and y=4mx has area a²/12. Use integration to find all possible nonzero real m.

✅ Step-by-Step Solution

Final Answer: m=±2

The line is x=y/(4m); the parabola is x=y²/(16a). Intersections have y=0 and y=4a/m.
First take m>0. Between y=0 and y=4a/m, the line is to the right of the parabola, so A=∫₀^(4a/m)[y/(4m)−y²/(16a)]dy.
Integrate: A=[y²/(8m)−y³/(48a)]₀^(4a/m)=2a²/(3m³).
For m<0 reflection in the x-axis gives the same area with |m|: A=2a²/(3|m|³). Equating to a²/12 gives |m|³=8.
Therefore |m|=2 and both m=2 and m=−2 satisfy the original area condition.

[1 mark] The line is x=y/(4m); the parabola is x=y²/(16a). Intersections have y=0 and y=4a/m.
[1 mark] First take m>0. Between y=0 and y=4a/m, the line is to the right of the parabola, so A=∫₀^(4a/m)[y/(4m)−y²/(16a)]dy.
[1 mark] Integrate: A=[y²/(8m)−y³/(48a)]₀^(4a/m)=2a²/(3m³).
[1 mark] For m<0 reflection in the x-axis gives the same area with |m|: A=2a²/(3|m|³). Equating to a²/12 gives |m|³=8.
[1 mark] Therefore |m|=2 and both m=2 and m=−2 satisfy the original area condition.

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