If y1,y2 and y3 are the ordinates of the vertices of a triangle inscribed in the parabola y2=4ax then its area is
A
12a|(y1−y2)(y2−y3)(y3−y1)|
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B
14a|(y1−y2)(y2−y3)(y3−y1)|
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C
18a|(y1−y2)(y2−y3)(y3−y1)|
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D
none
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Solution
The correct option is C18a|(y1−y2)(y2−y3)(y3−y1)| Vertices of the triangle are (y214a,y1)(y224a,y2)&(y234a,y3)
Area of the triangle = 12∣∣
∣∣14a(y21−y22)y1−y214a(y21−y22)y1−y3∣∣
∣∣=12(14a)|(y21−y22)(y1−y3)|
= 18a|(y1−y2)(y2−y3)(y3−y1)|