A) parabola
B) circle
C) hyperbola
D) ellipse
Correct Answer: C
Solution :
[c] Slope of tangent\[=\frac{dy}{dx}\] \[\therefore \]Slope of normal\[=-\frac{dx}{dy}\] Thus, the equation of normal is \[Y-y=-\frac{dx}{dy}(X-y)\] This meets x-axis (y=0), where \[-y=-\frac{dx}{dy}(X-x)\] or \[X=x+y\frac{dy}{dx}\] \[\therefore G\]is \[\left( x+y\frac{dy}{dx},0 \right)\] \[\therefore OG=2x\] \[\therefore x+y\frac{dy}{dx}=2x\] Or \[y\frac{dy}{dx}=x\] or \[ydy=xdx\] Integrating, we get \[\frac{{{y}^{2}}}{2}=\frac{{{x}^{2}}}{2}+\frac{C}{2}\] Or \[{{y}^{2}}-{{x}^{2}}=c,\]which is a hyperbola.
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