Conic Sections Question 172
Question: The ordinates of the triangle inscribed in parabola $ y^{2}=4ax $ are $ y_1,\ y_2,\ y_3 $ , then the area of triangle is
Options:
A) $ \frac{1}{8a}(y_1+y_2)(y_2+y_3)(y_3+y_1) $
B) $ \frac{1}{4a}(y_1+y_2)(y_2+y_3)(y_3+y_1) $
C) $ \frac{1}{8a}(y_1-y_2)(y_2-y_3)(y_3-y_1) $
D) $ \frac{1}{4a}(y_1-y_2)(y_2-y_3)(y_3-y_1) $
Show Answer
Answer:
Correct Answer: B
Solution:
Points $ ( \frac{y_1^{2}}{4a},y_1 ),( \frac{y_2^{2}}{4a},y_2 ),( \frac{y_3^{2}}{4a},y_3 ) $
Use area formula and get $ \Delta =\frac{1}{8a}(y_1-y_2)(y_2-y_3)(y_3-y_1) $ .
BETA
