Question

Which of the following points lies on the tangent to the curve ${\mathrm{x}}^4{\mathrm{e}}^{\mathrm{y}}+2\sqrt{(\mathrm{y}+1)}=3$ at the point $\left(1,0\right)$?

• A. $\left(2,6\right)$
• B. $\left(2,2\right)$
• C. $\left(-2,6\right)$
• D. $\left(-2,4\right)$

Answer 1

The correct option is C

$\left(-2,6\right)$

Explanation for the correct option:

The slope of the tangent to the curve is given as $\mathrm{f}\text{'}\left({\mathrm{x}}_0\right)$.

Step 1: Find the derivative of the given equation of the curve.

${\mathrm{x}}^4{\mathrm{e}}^{\mathrm{y}}+2\sqrt{(\mathrm{y}+1)}=3$

$\Rightarrow 4x^3{e}^y+x^4{e}^{y}y\text{'}+\frac{1}{\sqrt{\mathrm{y}+1}}\mathrm{y}\text{'}=0$

Step 2: Put the point$\left(1,0\right)$ in the derivative.

$$\begin{gathered} \Rightarrow 4e^0+e^0\mathrm{y}\text{'}+\frac{\mathrm{y}\text{'}}{\sqrt{1}}=0 \\ \Rightarrow 4+2\mathrm{y}\text{'}=0 \\ \Rightarrow \mathrm{y}\text{'}=-2 \end{gathered}$$

Step 3: Derive the equation of tangent.

We know that the equation of the tangent to any curve $y=f\left(x\right)$ is given by $y-y_0=f\text{'}\left(x_0\right)\left(x-x_0\right)$

The equation of the tangent

$$\begin{gathered} \mathrm{y}-0=-2\left(\mathrm{x}-1\right) \\ \Rightarrow y=-2x+2 \end{gathered}$$

Step 4: Put $\left(-2,6\right)$ in the equation of tangent.

$\mathrm{y}=-2\mathrm{x}+2$

$$\begin{gathered} \Rightarrow 6=-2\left(-2\right)+2 \\ \Rightarrow 6=6 \end{gathered}$$

Since the left-hand side is equal to the right-hand side, $\left(-2,6\right)$ is the point that lies on the tangent to the curve.

Hence, the correct answer is option C

Explanation of incorrect option:

Option (A)

Put $\left(2,6\right)$ in the equation of tangent.

$$\begin{gathered} \mathrm{y}=-2\mathrm{x}+2 \\ \Rightarrow 6=-2\left(2\right)+2 \\ \Rightarrow 6\ne -2 \end{gathered}$$

Since the left-hand side is not equal to the right-hand side, $\left(2,6\right)$ is not a point that lies on the tangent to the curve.

Option(B)

Put $\left(2,2\right)$ in the equation of tangent.

$$\begin{gathered} \mathrm{y}=-2\mathrm{x}+2 \\ \Rightarrow 2=-2\left(2\right)+2 \\ \Rightarrow 2\ne -2 \end{gathered}$$

Since the left-hand side is not equal to the right-hand side, $\left(2,2\right)$ is not a point that lies on the tangent to the curve.

Option(D)

Put $\left(-2,4\right)$ in the equation of tangent.

$\mathrm{y}=-2\mathrm{x}+2$

$\Rightarrow 4=-2\left(-2\right)+2$

$\Rightarrow 4\ne 6$

Since the left-hand side is not equal to the right-hand side, $\left(-2,4\right)$ is not a point that lies on the tangent to the curve.

So, the options A, B, and D are incorrect.

Hence, the correct answer is option C.