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# The curve given by x+y = ${{e}^{xy}}$ has a tangent parallel to the y-axis at the point-(a) (0, 1)(b) (1, 0)(c) (1, 1)(d) (-1, -1)

Last updated date: 17th Jul 2024
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Hint: To solve the problem, we should be aware about the formula of tangent to the curve. The formula of tangent to a curve is given by $\dfrac{dy}{dx}$.
Further, for a tangent to be parallel to the y-axis, the slope should be infinity.

Complete step by step solution:
Before we begin to solve the problem, we should know briefly about a tangent to the curve. Basically, a tangent to the curve is a straight line that touches a curve but does not cross it. Thus, to find the tangent at point (x,y), we find $\dfrac{dy}{dx}$ of the curve and then put points (x,y) in the expression. This will be illustrated by solving the above problem.
We have,
x+y = ${{e}^{xy}}$ -- (1)
Now, performing derivative with respect to x on (1). We get,
$\dfrac{d}{dx}(x+y)=\dfrac{d}{dx}({{e}^{xy}})$
1+$\dfrac{dy}{dx}$=${{e}^{xy}}\dfrac{d}{dx}(xy)$
Now, to evaluate $\dfrac{d}{dx}(xy)$, we utilize the product rule to solve this.
According to the product rule, for two functions f(x) and g(x), we have,
$\dfrac{d}{dx}\left( f(x)g(x) \right)=\left[ \dfrac{d}{dx}(f(x)) \right]g(x)+\left[ \dfrac{d}{dx}(g(x)) \right]f(x)$
We will utilize this to solve the below differentiation of xy. Thus, we have,
1+$\dfrac{dy}{dx}$=${{e}^{xy}}\left( x\dfrac{dy}{dx}+y \right)$
$\dfrac{dy}{dx}\left( 1-x{{e}^{xy}} \right)=y{{e}^{xy}}-1$
$\dfrac{dy}{dx}=\dfrac{y{{e}^{xy}}-1}{\left( 1-x{{e}^{xy}} \right)}$
This derivative is evaluated at points (x,y) on the curve x+y = ${{e}^{xy}}$. This is the slope of the curve at point (x,y).
Now, to ensure that the tangent is parallel to the y-axis. For this condition to hold true, the slope of the line should be infinity. Thus, the denominator of slope should be zero. Thus, we get,
$x{{e}^{xy}}$=1 -- (2)
Now, we start putting in options given in the problem. We see that only (1,0) satisfies equation (2).

Note: An alternative approach to solving the above problem would be to find the normal to the curve instead of the tangent. Since, normal to the curve is perpendicular to the tangent at that point. Thus, normal at the point (for the tangent to be parallel to the y-axis) would be 0. The normal to the curve is-
$-\dfrac{dx}{dy}=\dfrac{\left( 1-x{{e}^{xy}} \right)}{y{{e}^{xy}}-1}$
Thus, this should be 0. Similar to the solution, we would again have $x{{e}^{xy}}$=1.