Question
If the tangent $(1,1)$ on $y^{2}=x(2-x)^{2}$ meets the curve again at $P$, then $P$ is(a) $(-1,2)$(b) $(4,4)$(c) $\left(\frac{9}{4}, \frac{3}{8}\right)$(d) None
Step 1
This gives us $2y\frac{dy}{dx} = 2(2-x)^2 - 2x(2-x)$. Show more…
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