-3
$\begingroup$

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

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

$\endgroup$

closed as off-topic by TastyRomeo, Namaste, Leucippus, Henrik, user223391 Jan 14 '17 at 18:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TastyRomeo, Namaste, Leucippus, Henrik, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Hi and Welcome to MSE, Take a tour and read some questions and answers. These things will help you in understanding the working of site in a better way. $\endgroup$ – Harsh Kumar Jan 13 '17 at 15:57
2
$\begingroup$

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!!!

$\endgroup$
3
$\begingroup$

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 $$t>0$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.