JEE Advance - Mathematics (2019 - Paper 1 Offline - No. 5)
Let $$\Gamma $$ denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to I` at a point P intersect the y-axis at YP. If PYP has length 1 for each point P on I`, then which of the following options is/are correct?
$$xy' + \sqrt {1 - {x^2}} = 0$$
$$xy' - \sqrt {1 - {x^2}} = 0$$
$$y = {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $$
$$y = - {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} $$
Explanation
Let a point P(h, k) on the curve y = y(x), so equation of tangent to the curve at point P is
$$y - k = {\left( {{{dy} \over {dx}}} \right)_{h,k}}(x - h)$$ ....(i)
Now, the tangent (i) intersect the Y-axis at Yp, so coordinates Yp is $$\left( {0,k - h{{dy} \over {dx}}} \right)$$,
where $${{{dy} \over {dx}}}$$ = $${\left( {{{dy} \over {dx}}} \right)}$$(h,k)
So, PYp = 1 (given)
$$ \Rightarrow \sqrt {{h^2} + {h^2}{{\left( {{{dy} \over {dx}}} \right)}^2}} = 1$$
$$ \Rightarrow {{dy} \over {dx}} = \pm {{\sqrt {1 - {x^2}} } \over x}$$
[on replacing h by x]
$$ \Rightarrow dy = \pm {{\sqrt {1 - {x^2}} } \over x}dx$$
On puting x = sin$$\theta $$, dx = cos$$\theta $$d$$\theta $$, we get
$$dy = \pm {{\sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}\cos \theta d\theta $$
$$ = \pm {{{{\cos }^2}\theta } \over {\sin \theta }}d\theta $$
$$ = \pm (\cos ec\theta - \sin \theta )d\theta $$
$$ \Rightarrow y = \pm [1n(\cos ec\theta - \cot \theta ) + \cos \theta ] + C$$
$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) + \cos \theta } \right] + C$$
$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}} \right) + \sqrt {1 - {{\sin }^2}\theta } } \right] + C$$
$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} } \right] + C$$
[$$ \because $$ x = sin$$\theta $$]
$$ = \pm \left[ { - 1n{{1 + \sqrt {1 - {x^2}} } \over x} + \sqrt {1 - {x^2}} } \right] + C$$
[on rationalization]
$$ \because $$ The curve is in the first quadrant so y must be positive, so
$$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} + C$$
As curve passes through (1, 0), so
$$0 = 0 - 0 + c \Rightarrow c = 0$$, so required curve is
$$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $$
and required differential eaquation is
$${{dy} \over {dx}} = - {{\sqrt {1 - {x^2}} } \over x}$$
$$ \Rightarrow xy' + \sqrt {1 - {x^2}} = 0$$
Hence, options (a) and (c) are correct.
$$y - k = {\left( {{{dy} \over {dx}}} \right)_{h,k}}(x - h)$$ ....(i)
Now, the tangent (i) intersect the Y-axis at Yp, so coordinates Yp is $$\left( {0,k - h{{dy} \over {dx}}} \right)$$,
where $${{{dy} \over {dx}}}$$ = $${\left( {{{dy} \over {dx}}} \right)}$$(h,k)
So, PYp = 1 (given)
$$ \Rightarrow \sqrt {{h^2} + {h^2}{{\left( {{{dy} \over {dx}}} \right)}^2}} = 1$$
$$ \Rightarrow {{dy} \over {dx}} = \pm {{\sqrt {1 - {x^2}} } \over x}$$
[on replacing h by x]
$$ \Rightarrow dy = \pm {{\sqrt {1 - {x^2}} } \over x}dx$$
On puting x = sin$$\theta $$, dx = cos$$\theta $$d$$\theta $$, we get
$$dy = \pm {{\sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}\cos \theta d\theta $$
$$ = \pm {{{{\cos }^2}\theta } \over {\sin \theta }}d\theta $$
$$ = \pm (\cos ec\theta - \sin \theta )d\theta $$
$$ \Rightarrow y = \pm [1n(\cos ec\theta - \cot \theta ) + \cos \theta ] + C$$
$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) + \cos \theta } \right] + C$$
$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}} \right) + \sqrt {1 - {{\sin }^2}\theta } } \right] + C$$
$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} } \right] + C$$
[$$ \because $$ x = sin$$\theta $$]
$$ = \pm \left[ { - 1n{{1 + \sqrt {1 - {x^2}} } \over x} + \sqrt {1 - {x^2}} } \right] + C$$
[on rationalization]
$$ \because $$ The curve is in the first quadrant so y must be positive, so
$$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} + C$$
As curve passes through (1, 0), so
$$0 = 0 - 0 + c \Rightarrow c = 0$$, so required curve is
$$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $$
and required differential eaquation is
$${{dy} \over {dx}} = - {{\sqrt {1 - {x^2}} } \over x}$$
$$ \Rightarrow xy' + \sqrt {1 - {x^2}} = 0$$
Hence, options (a) and (c) are correct.
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