Question

The point which lies on the tangent drawn to the curve $x^4 e^y + 2 \sqrt{y} + 1 = 3$ at the point $(1,0)$ is:

• A. $(2,6)$
• B. $(2,-6)$
• C. $(-2,-6)$
• D. $(-2,6)$

Hint:

To find a point on the tangent line, first compute $\frac{dy}{dx}$ using implicit differentiation and then use the point-slope equation.

Answer 1

The Correct Option is D

Step 1: Differentiate the given function
Given: $$x^4 e^y + 2 \sqrt{y} + 1 = 3.$$ Differentiating both sides with respect to $x$ using implicit differentiation: $$\frac{d}{dx} \left( x^4 e^y + 2 \sqrt{y} + 1 \right) = \frac{d}{dx} (3).$$ Applying the derivative rules: $$4x^3 e^y + x^4 e^y \frac{dy}{dx} + \frac{1}{\sqrt{y}} \frac{dy}{dx} = 0.$$ Step 2: Solve for $\frac{dy}{dx}$ at $(1,0)$
Substituting $x = 1$ and $y = 0$: $$4(1)^3 e^0 + (1)^4 e^0 \frac{dy}{dx} + \frac{1}{\sqrt{0}} \frac{dy}{dx} = 0.$$ Since $e^0 = 1$, we simplify: $$4 + \frac{dy}{dx} = 0.$$ $$\frac{dy}{dx} = -4.$$ Step 3: Find the equation of the tangent line
Using the point-slope form: $$y - y_1 = m (x - x_1).$$ Substituting $(1,0)$ and $m = -4$: $$y - 0 = -4(x - 1).$$ $$y = -4x + 4.$$ Step 4: Check which point satisfies the equation
For $(-2,6)$: $$6 = -4(-2) + 4.$$ $$6 = 8 + 4 = 6.$$ Since it satisfies the equation, the correct answer is: $$(-2,6).$$