A) \[\frac{15{{a}^{3}}}{4}\]sq units
B) \[\frac{15{{a}^{2}}}{4}\]sq units
C) \[\frac{7{{a}^{2}}}{4}\]sq units
D) \[\frac{{{a}^{2}}}{4}\]sq units
Correct Answer: B
Solution :
Equation of any tangent of the parabola,\[{{y}^{2}}=4ax\]is\[y=mx+\frac{a}{m}\]. This line will touch the circle\[{{x}^{2}}+{{y}^{2}}=\frac{{{a}^{2}}}{2}\]. \[\therefore \] \[BC=2\sqrt{O{{B}^{2}}-O{{K}^{2}}}\] \[=2\sqrt{\frac{{{a}^{2}}}{2}-\frac{{{a}^{2}}}{4}}\] and we know that DE is the latusrectum of the parabola, so its length is\[4a\]. Thus, area of trapezium\[BCDE\] \[=\frac{1}{2}(BC+DE)(KL)\] \[=\frac{1}{2}(a+4a)\left[ \frac{3a}{2} \right]\] \[=\frac{15{{a}^{2}}}{4}\,\,sq\,\,units\]You need to login to perform this action.
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