question_answer

From a point A common tangents are drawn to the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}/2\]and parabola\[{{y}^{2}}=4ax\]. The area of the quadrilateral formed by the common tangents to the chord of contact of the parabola is

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\[BCDE\]                 \[=\frac{1}{2}(BC+DE)(KL)\]                 \[=\frac{1}{2}(a+4a)\left[ \frac{3a}{2} \right]\]                 \[=\frac{15{{a}^{2}}}{4}\,\,sq\,\,units\]