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Q. If the tangent at $(1,1)$ on $y^2=x(2-x)^2$ meets the curve again at $P$ then the co-ordinates of $P$ are

Application of Derivatives

Solution:

$\left.\frac{ dy }{ dx }\right|_{(1,1)}=-\frac{1}{2} ; \quad \frac{ y _1-1}{ x _1-1}=-\frac{1}{2} ; 2 y _1+ x _1=3$ which is satisfying option (1)