A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with :
Solution
<p>According to the question (Let P(x, y))</p>
<p>$2x - y{{dx} \over {dy}} = 0$</p>
<p>($\because$ equation of tangent at $P:y - y = {{dy} \over {dx}}(y - x)$)</p>
<p>$\therefore$ $2{{dy} \over {y}} = {{dx} \over x}$</p>
<p>$\Rightarrow 2\ln y = \ln x + \ln c$</p>
<p>$\Rightarrow {y^2} = cx$</p>
<p>$\because$ this curve passes through (3, 3)</p>
<p>$\therefore$ c = 3</p>
<p>$\therefore$ required parabola ${y^2} = 3x$ and L.R = 3</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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