Question

The curve $y - {e^{xy}} + x = 0$ has a vertical tangent at the point-
A.$\left( {1,1} \right)$
B. No point.
C. $\left( {0,1} \right)$
D.$\left( {1,0} \right)$

Answer 1

First differentiate the given equation with respect to x. Solve the differential equation and put$\Rightarrow \dfrac{{dx}}{{dy}} = 0$ as the function having a vertical tangent is not differentiable at the point of tangency because its slope is${90^ \circ }$. Put this value in the differential equation and solve it. Then substitute the value of ${e^{xy}}$ in the equation and solve for x. The values you get will be the point at which the tangent lies.

Complete step by step answer:

Given curve-$y - {e^{xy}} + x = 0$-- (i)
We can also write ${e^{xy}} = y + x$ --- (ii)
On differentiating eq. (i) with respect to x, we get
$\Rightarrow \dfrac{d}{{dx}}\left( {y - {e^{xy}} + x} \right) = 0$
$\Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{d}{{dx}}{e^{xy}} + \dfrac{d}{{dx}}x = 0$
On using chain rule,
$\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right).g'\left( x \right)$
We get,
$\Rightarrow \dfrac{{dy}}{{dx}} - {e^{xy}}\dfrac{d}{{dx}}\left( {xy} \right) + 1 = 0$
On using product rule, we get-
$\dfrac{d}{{dx}}\left[ {f\left( x \right).g\left( x \right)} \right] = f\left( x \right).g'\left( x \right) + g\left( x \right)f'\left( x \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} - {e^{xy}}\left( {x\dfrac{{dy}}{{dx}} + y.1} \right) + 1 = 0$
On simplifying we get,
$\Rightarrow \dfrac{{dy}}{{dx}} - {e^{xy}}.x\dfrac{{dy}}{{dx}} - {e^{xy}}.y + 1 = 0$
On taking first derivative common we get,
$\Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - x{e^{xy}}} \right) - y{e^{xy}} + 1 = 0$
$\Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - x{e^{xy}}} \right) = y{e^{xy}} - 1$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y{e^{xy}} - 1}}{{\left( {1 - x{e^{xy}}} \right)}}$
Since we know that derivative of function at a vertical tangent is not defined as the slope of tangent is ${90^ \circ }$ then,
$\Rightarrow \dfrac{{dy}}{{dx}} = \infty$
Then
$\Rightarrow \dfrac{{dx}}{{dy}} = 0$
On putting the values we get,
$\Rightarrow \dfrac{{\left( {1 - x{e^{xy}}} \right)}}{{y{e^{xy}} - 1}} = 0$
Hence we get,
$\Rightarrow$ $1 - x{e^{xy}} = 0$
$\Rightarrow x{e^{xy}} = 1$
Substituting value from eq. (ii), we get-
$\Rightarrow x\left( {y + x} \right) = 1$
On multiplying we get,
$\Rightarrow xy + {x^2} = 1$
On adjusting the equation we get,
$\Rightarrow xy = 1 - {x^2}$
Here we can find the value of y.
$\Rightarrow y = \dfrac{{1 - {x^2}}}{x}$
On putting y=$0$ then we get,
$\Rightarrow {x^2} - 1 = 0$
$\Rightarrow {x^2} = 1$
On removing the square, we get-
$\Rightarrow x = \sqrt 1$
$\Rightarrow x = \pm 1$
So we get two points $\left( {1,0} \right)$ and$\left( { - 1,0} \right)$ . But $\left( { - 1,0} \right)$ does not lie on the curve only point $\left( {1,0} \right)$ lies on the curve.
Hence the given curve has vertical tangent at point$\left( {1,0} \right)$.
Answer- The correct answer is D.

Note: Here students may get confused how $\dfrac{{dx}}{{dy}} = 0$ . A vertical tangent is a line that is vertical and a vertical line has infinite slopes. So since tangent is vertical then the slope is ${90^ \circ }$ which means that slope m=$\tan {90^ \circ }$ .And we know that $\dfrac{{dy}}{{dx}}$ is the slope of the given curve hence we can write,
$\Rightarrow \dfrac{{dy}}{{dx}} = \tan {90^ \circ } = \infty$ so we can also write-$\dfrac{{dx}}{{dy}} = 0$. So we can say that the derivative is undefined at a vertical tangent.