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The line $2 y=3 x+12$ cuts the parabola $4 y=3 x^{2}.$
Where does the line cut the parabola?
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Where does the line cut the parabola?
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The correct answer is:
At both $(-2,3)$ and $(4,12)$
Equation of line $2 y=3 x+12$
Equation of parabola $4 y=3 x^{2}$
From eqs. (i) and (ii), we get $2(3 x+12)=3 x^{2}$
$3 x^{2}-6 x-24=0$
$x^{2}-2 x-8=0$
$(x-4)(x+2)=0$
$\therefore \mathrm{x}=4$
and $x=-2$ Now putting the value of $\mathrm{x}$ in eqn (ii) We get $y=12$ and $y=3$ Thus, the points $(-2,3)$ and $(4,12)$
Equation of parabola $4 y=3 x^{2}$
From eqs. (i) and (ii), we get $2(3 x+12)=3 x^{2}$
$3 x^{2}-6 x-24=0$
$x^{2}-2 x-8=0$
$(x-4)(x+2)=0$
$\therefore \mathrm{x}=4$
and $x=-2$ Now putting the value of $\mathrm{x}$ in eqn (ii) We get $y=12$ and $y=3$ Thus, the points $(-2,3)$ and $(4,12)$
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