Question

The line $2y=3x+12$ cuts the parabola $4y=3x^2$. What is the area enclosed by the parabola, the line and the y-axis in the first quadrant?

• A. $7$ square unit
• B. $14$ square unit
• C. $20$ square unit
• D. $21$ square unit

Answer 1

The correct option is C $20$ square unit

The graph is shown in the image.

The intersection point of $2y=3x+12$ and $4y=3x^2$

$2(\frac{3x^2}{4})=3x+12$

$x^2-2x-8=0$

$x=-2,4$

Hence, intersection point are $(-2,3)$ and $(4,12)$

Now area bounded by given curve, line and the $Y$-axis in 1^{st} quadrant

Now $A={\int }_0^4[\frac{3x+12}{2}]-{\int }_0^4[\frac{3x^2}{4}]$

$A=[\frac{3x^2}{4}+6x]{|}_0^4-[\frac{x^3}{4}]{|}_0^4$

$A=12+24-16$

$A=20$ sq.unit