Question

The line 2y=3x+12 cuts the parabola 4y=3x2. Where does the line cut the parabola?

• A. At $(-2,3)$ only
• B. At $(4,12)$ only
• C. At both $(-2,3)$ and $(4,12)$
• D. Neither at $(-2,3)$ nor at $(4,12)$

Answer 1

The correct option is C At both $(-2,3)$ and $(4,12)$

Given f(x) is $2y=3x+12\Rightarrow y=\frac{3x+12}{2}......\left(1\right)$

and g(x) is $4y=3x^2\Rightarrow y=\frac{3x^2}{4}.......\left(2\right)$

By equating equ(1) and equ(2) we get

$4(3x+12)=2\left(3x^2\right)$ [dividing by 2 on both sides]

$\Rightarrow 2(3x+12)=3x^2$

$\Rightarrow 6x+24=3x^2$

$\therefore 3x^2-6x-24=0$

dividing through out by $3$ we get

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

On factorising we get $(x-4)(x+2)$ or $x=4,-2$;hence $y=3,12$ respectively.

The points of intersection are $(-2,3)$ and $(4,12)$.