The normal to a curve at P(x,y) meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a :
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Official solution
Equation of normal at \[P\left( {x,y} \right)\] is \[Y - y = - {{dx} \over {dy}}\left( {x - x} \right)\]
Coordinate of G at X axis is \[\left( {X,0} \right)\] (let)
\[\therefore\] \[0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x\]
\[ \Rightarrow X = x + y{{dy} \over {dx}}\] \[\therefore\] Co-ordinate of \[G\left( {x + y{{dy} \over {dx}},0} \right)\]
Given distance of G from origin = twice of the abscissa of P.
as distance cannot be −ve, therefore abscissa x should be +ve
\[\therefore\] \[x + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx\]
On Integrating \[ \Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}\]
\[\therefore\] the curve is a hyperbola