Question

The line $2y=3x+12$ cuts the parabola $4y=3x^2$. What is the area enclosed by the parabola and the line?

• A. $27$ square unit
• B. $36$ square unit
• C. $48$ square unit
• D. $54$ square unit

Answer 1

The correct option is A $27$ 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, $A={\int }_{-2}^4[\frac{3x+12}{2}]-{\int }_{-2}^4[\frac{3x^2}{4}]$

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

$A=9+36-18$

$A=27$ $sq.unit$