Question

The angle at which the curve $y=a{e}^{ax}$ intersects the $y-\text{axis}$ is

• A. ${\tan }^{-1}\left(a^2\right)$
• B. ${\sin }^{-1}\left(\frac{1}{\sqrt{1+a^4}}\right)$
• C. ${\text{cot}}^{-1}\left(a^2\right)$
• D. ${\sin }^{-1}\left(\sqrt{1+a^4}\right)$

Answer 1

The correct option is C

${\text{cot}}^{-1}\left(a^2\right)$

Find angle at which the curve$y=a{e}^{ax}$ intersects the $y-\text{axis}$

Consider the curve $y=a{e}^{ax}$

Slope of the tangent is expressed as $m=\frac{dy}{dx}$

Then,

$$\begin{gathered} m=a^2{e}^{ax} \\ m=a^2\left[\text{At}x=0\right] \end{gathered}$$

The curve intersects $y-\text{axis}$ at $\left(0,a\right)$.

Consider,

$$\begin{gathered} m_1=a^2 \\ {m}_2=\infty \end{gathered}$$

Let $\theta$ be an angle of two curve

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Rewrite the above Equation as,

$$\begin{gathered} \tan \theta =\left|\frac{{\displaystyle \frac{m_1}{m_2}}-1}{{\displaystyle \frac{1}{m_2}}+m_1}\right| \\ \tan \theta =\left|\frac{{\displaystyle \frac{a^2}{\infty }-1}}{{\displaystyle \frac{1}{\infty }+a^2}}\right| \\ \tan \theta =\left|\frac{{\displaystyle -1}}{{\displaystyle a^2}}\right| \\ \tan \theta =\frac{{\displaystyle 1}}{{\displaystyle a^2}} \\ \theta ={\tan }^{-1}\frac{{\displaystyle -1}}{{\displaystyle a^2}} \\ \theta =co{t}^{-1}{a}^2 \end{gathered}$$

Hence, the correct answer is Option C.