• question_answer A curve passing through M(1, 1) have the property that any tangent intersects the y-axis at the point which is equidistant from the point of tangency and the origin, can be A)  straight line                   B)  circle C)  parabola                       D)  hyperbola

$Y-y\,=m(X-x)$ Put $X=0$ $Y=y-mx$ Hence $A\equiv \,(0,\,\,y-mx)$ $\therefore \,\,OA=OP$ $\Rightarrow \,\,|y-mx{{|}^{2}}={{x}^{2}}+{{m}^{2}}{{x}^{2}}$ $\Rightarrow \,\,{{y}^{2}}+{{m}^{2}}{{x}^{2}}-2xym\,={{x}^{2}}\,+{{m}^{2}}{{x}^{2}}$ $\therefore \,\frac{dy}{dx}\,=\frac{{{y}^{2}}-{{x}^{2}}}{2xy}\,\Rightarrow \,\,2xy\,\frac{dy}{dx}\,={{y}^{2}}-{{x}^{2}}$ Now put ${{y}^{2}}=t\Rightarrow \,2y\,\frac{dy}{dx}\,=\frac{dt}{dx}$ $\therefore \,\,x\frac{dt}{dx}=t-{{x}^{2}};\,\,\frac{dt}{dx}-\frac{1}{x}\,t=-x$ Integrating factor $={{e}^{-\frac{1}{x}dx}}\,={{e}^{-\ln x}}=\frac{1}{x};$ Hence $\frac{1}{x}.t=-\int_{{}}^{{}}{dx;\,\frac{t}{x}\,=-x+C}$ $\therefore \,\,{{y}^{2}}=-{{x}^{2}}+Cx & ;$ $\Rightarrow \,{{x}^{2}}+{{y}^{2}}\,=Cx\Rightarrow \,$ Circle.