The straight line $y=tx-2$ is a tangent to the graph of a curve $y=2x^2 +4x$.

Find the value of $t$ ($t>0$)


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HINT: Solve the equation of tangent with the equation of curve. Then you will have a quadratic equation solve it for $Discriminant=0$

You have the equations $y=tx-2\tag1$ and $y=2x^2+4x\tag2$ put the value of $y$ from equation $(1)$ to equation $(2)$ then you will have $$tx-2=2x^2+4x$$ $$2x^2+x(4-t)+2=0$$ since the line is tangent hence equation $(3)$ has equal roots.
$\therefore$ $$(4-t)^2-4(2)(2)=0$$ $$t^2-8t=0$$ from here it is clear that $t=0$ or $t=8$ which means $$t=8$$ since $t>0$

Hope it helps!!!


solving the equation $$tx-2=2x^2+4x$$ we get $$x_{1,2}=\frac{1}{4}t-1\pm\frac{1}{4}\sqrt{t^2-8t}$$ for a tangent line must the discriminat equal to zero, thus we have $$t=8$$ since $$t>0$$


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