The curve given by x + y = ${e}^{xy}$ has to tangent parallel to the y - axis at that point.
A.(0, 1)
B.(1, 0)
C.(1, 1)
D.none of these
Answer
640.2k+ views
Hint: Differentiate the curve with respect to x to find slope and check the option/ point which satisfies the condition(slope of line parallel to y-axis is undefined).
Complete step-by-step answer:
To find the tangent we have to differentiate the equation first.
x + y = ${e}^{xy}$
On differentiating the term
$1 + \dfrac{{dy}}{{dx}} = {e^{xy}}\left( {x\dfrac{{dy}}{{dx}} + y} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y{e^{xy}} - 1}}{{1 - x{e^{xy}}}}$
Now checking by option,
${\left| {\dfrac{{dy}}{{dx}}} \right|_{(0,1)}} = \dfrac{{1 - 1}}{{1 - 0}} = \dfrac{0}{1} = 0$ not coming $\infty $
${\left| {\dfrac{{dy}}{{dx}}} \right|_{(1,0)}} = \dfrac{{ - 1}}{0} = \infty $
∴ This point satisfies our condition.
Hence option B (1, 0) is correct.
Note: Slope of a line parallel to y-axis is infinite and slope of a line parallel to x-axis is zero. Slope of the curve is defined by the differentiation of the curve with respect to x.
Complete step-by-step answer:
To find the tangent we have to differentiate the equation first.
x + y = ${e}^{xy}$
On differentiating the term
$1 + \dfrac{{dy}}{{dx}} = {e^{xy}}\left( {x\dfrac{{dy}}{{dx}} + y} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y{e^{xy}} - 1}}{{1 - x{e^{xy}}}}$
Now checking by option,
${\left| {\dfrac{{dy}}{{dx}}} \right|_{(0,1)}} = \dfrac{{1 - 1}}{{1 - 0}} = \dfrac{0}{1} = 0$ not coming $\infty $
${\left| {\dfrac{{dy}}{{dx}}} \right|_{(1,0)}} = \dfrac{{ - 1}}{0} = \infty $
∴ This point satisfies our condition.
Hence option B (1, 0) is correct.
Note: Slope of a line parallel to y-axis is infinite and slope of a line parallel to x-axis is zero. Slope of the curve is defined by the differentiation of the curve with respect to x.
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