Question

If the area of the triangle inscribed in the parabola $y^2=8x$, then the ordinate of whose vertices are $1,3$ and $4$ is

• A. $\frac{3}{4}$
• B. $\frac{3}{8}$
• C. $\frac{4}{3}$
• D. $\frac{5}{4}$

Answer 1

Answer: (B)

$\frac{3}{8}$

Given,

$y^2=8x$

$\Rightarrow a=2$

$y_1=1,y_2=3,y_3=4$

we have formula,

area of triangle inscribed in a parabola is,

$=\frac{(y_1-y_2)(y_2-y_3)(y_3-y_1)}{8a}$

$=\frac{(1-3)(3-4)(4-1)}{8\times 2}$

$=\frac{6}{16}$

$=\frac{3}{8}$