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: 28th Mar 2023
•
Total views: 306.6k
•
Views today: 5.83k
Answer
306.6k+ views
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.
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Ray optics is valid when characteristic dimensions class 12 physics CBSE

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

Alfred Wallace worked in A Galapagos Island B Australian class 12 biology CBSE

Imagine an atom made up of a proton and a hypothetical class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Why is the cell called the structural and functional class 12 biology CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main
