JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 7)
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 :
length of latus rectum 3
length of latus rectum 6
focus $$\left( {{4 \over 3},0} \right)$$
focus $$\left( {0,{3 \over 4}} \right)$$
Explanation
According to the question (Let P(x, y))
$$2x - y{{dx} \over {dy}} = 0$$
($$\because$$ equation of tangent at $$P:y - y = {{dy} \over {dx}}(y - x)$$)
$$\therefore$$ $$2{{dy} \over {y}} = {{dx} \over x}$$
$$ \Rightarrow 2\ln y = \ln x + \ln c$$
$$ \Rightarrow {y^2} = cx$$
$$\because$$ this curve passes through (3, 3)
$$\therefore$$ c = 3
$$\therefore$$ required parabola $${y^2} = 3x$$ and L.R = 3
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