A) 6
B) 9
C) 12
D) 3
Correct Answer: B
Solution :
[b] The given parabola is \[{{(y-2)}^{2}}=x-1\] |
Vertex (1, 2) and it meets x - axis at (5, 0) |
Also it gives \[{{y}^{2}}-4y-x+5=0\] |
So, that equation of tangent to the parabola at (2, 3) is |
\[y.3-2(y+3)-\frac{1}{2}(x+2)+5=0\] or \[x-2y+4=0\] |
which meets x-axis at (-4, 0). |
In the figure shaded area is the required area. |
Let us draw PD perpendicular to y - axis. |
Then required area \[=Ar\,\,\Delta BOA+Ar(OCPD)\] |
\[-Ar(\Delta APD)\] |
\[=\frac{1}{2}\times 4\times 2+\int_{0}^{3}{xdy-\frac{1}{2}\times 2\times 1=}\] |
\[3+\int_{0}^{3}{{{(y-2)}^{2}}+1dy}\] |
\[=3+\left[ \frac{{{(y-2)}^{3}}}{3}+y \right]_{0}^{3}=3+\left[ \frac{1}{3}+3+\frac{8}{3} \right]\] |
\[=3+6=9\] Sq. units |
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