For a > 0, let the curves C1 : y2 = ax and
C2 : x2 = ay intersect at origin O and a point P.
Let the line x = b (0 < b < a) intersect the chord
OP and the x-axis at points Q and R,
respectively. If the line x = b bisects the area
bounded by the curves, C1 and C2, and the area of
\[\Delta \]OQR = \[{1 \over 2}\], then 'a' satisfies the equation :
Solutions · 1
Official solution
C1 : y2
= ax, C2 : x2
= ay (a > 0)
P : intersection point of y2
= ax and x2
= ay
\[ \Rightarrow \] x4
= a2
y2
\[ \Rightarrow \] x4
= a2ax
\[ \Rightarrow \] x = a, y = a
\[ \therefore \] Point P : (a, a)
Line OP : y = x
\[ \Rightarrow \] Point Q = (b, b)
Area \[\Delta \]OQR = \[{1 \over 2}\]
\[ \Rightarrow \] \[{1 \over 2} \times b \times b\] = \[{1 \over 2}\]
\[ \Rightarrow \] b = 1
As line x = b bisect the area between curve
\[ \therefore \] \[{1 \over 2}\int\limits_0^a {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx\] = \[\int\limits_0^1 {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx\]
\[ \Rightarrow \] \[\left[ {{1 \over 2}\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {1 \over 2}{{{x^3}} \over {3a}}} \right]_0^a\] = \[\left[ {\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {{{x^3}} \over {3a}}} \right]_0^1\]
\[ \Rightarrow \] \[{{{a^2}} \over 3} - {1 \over 2}{{{a^2}} \over 3}\] - 0 = \[{{2\sqrt a } \over 3} - {1 \over {3a}}\] - 0
\[ \Rightarrow \] \[{{{a^2}} \over 2} + {1 \over a} = 2\sqrt a \]
\[ \Rightarrow \] \[{a^3} + 2 = 4a\sqrt a \]
Squareing both sides, we get
\[ \Rightarrow \] a6+4a3+4=16a3
\[ \Rightarrow \] a6−12a3+4= 0
Hence a satisfy x6 -- 12x3 + 4 = 0.