c=-1

2 : 1

(i) not exist sq.units.
(ii) \(\frac { 32 }{ 15 } \)

Given sides are y = 2x + 1.....(1)
y = 3x + 1...(2)
x = 4...(3)
Solving (1) & (2), x = 0, y = 1
Solving (2) & (3), x = 4, y = 13
Solving (1) & (3), x = 4, y = 9

∴ Required area \(\int _{ 0 }^{ 4 }{ (3x+1) } dx-\int _{ 0 }^{ 4 }{ (2x+1) } dx\)
\({ =\left( \frac { { 3x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }-{ \left( \frac { { 2x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }\)
\(=\left( \frac { 48 }{ 2 } +4 \right) -(16+4)=28-20\)
Area = 8 sq. units

y = 2at ⇒ \(t=\frac { y }{ 2a } \) ...(1)
x = at² ... (2)
substituting (1) in (2) we get

\(x=a\left( \frac { { y }^{ 2 } }{ 4{ a }^{ 2 } } \right) =\frac { { y }^{ 2 } }{ 4a } \)
y² = 4ax
since y² = 4ax is symmetrical about x - axis,
Required area = \(2\int _{ 0 }^{ 4a }{ ydx } =2\int _{ 0 }^{ 4a }{ \sqrt { 4a \ x \ } } dx\)
\(=4\sqrt { a } \int _{ 0 }^{ 4a }{ { \sqrt { x } dx=4\sqrt { a } \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ a }^{ 4a } } \)
\(=\frac { 8\sqrt { a } }{ 3 } [4a\sqrt { 4a } -a\sqrt { a } ]\)
\(=\frac { 8\sqrt { a } }{ 3 } [8a\sqrt { a } -a\sqrt { a } ]\)
\(=\frac { 8\sqrt { a } }{ 3 } \times 7a\sqrt { a } \)
Area = \(\frac { { 56 }a^{ 2 } }{ 3 } \) sq.units