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The vertices of a triangle ABC are A(1,0) while B and C lie on the parabola ${ y = 2x - x^2}$. If AB = AC = $\frac{\sqrt{7}}{3}$ , then it's area in sq. units is.………

I plotted the parabola and it came to be symmetrical along the line $x $$= 1$ with its vertex at $(1,1)$. I tried to solve the question but could not approach it properly. Can someone help?

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  • $\begingroup$ ?The question is to find the area? $\endgroup$ – Juniven Jan 8 '17 at 3:00
  • $\begingroup$ Yes area is to be determined $\endgroup$ – Resorcinol Jan 8 '17 at 3:02
  • $\begingroup$ I hope you won't mind. Please encode the image next time if you have similar question like this. This is for the benefit of all the readers. $\endgroup$ – Juniven Jan 8 '17 at 3:04
  • $\begingroup$ Actually I don't know how to encode the image. It would be helpful if someone make the necessary edit. Thanks for you r suggestion. $\endgroup$ – Resorcinol Jan 8 '17 at 3:07
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    $\begingroup$ You could follow this guide and do it yourself. $\endgroup$ – Arthur Jan 8 '17 at 3:10
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Hint: find point $B(x,y)$ by solving the system:

$$ \begin{cases} \begin{align} y & = 2x-x^2 \\ (x-1)^2 + y^2 & = \frac{7}{9} \end{align} \end{cases} $$

For a shortcut, note that the first equation can be rewritten as $y=-(x-1)^2+1$, then eliminating the $(x-1)^2$ term between the two equations gives a simple quadratic in $y$.

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