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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 :

JEE · Math · previous-year question

  1. A.circle
  2. B.hyperbolacorrect
  3. C.ellipse
  4. D.parabola

Answer

B. hyperbola

Explanation

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

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