Question

If the point on $\mathrm{y}=\mathrm{xtan}\left(\alpha \right)-\left(\frac{{\mathrm{ax}}^2}{2{\mathrm{u}}^2{\cos }^2\left(\alpha \right)}\right)(\alpha >0)$ where the tangent is parallel to $y=x$ has an ordinate $\frac{u^2}{4a}$, then $a$ is equal to

• A. $\frac{\mathrm{\pi }}{6}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{3}$

Step 1. Find the value of $x$

Given $\mathrm{y}=\mathrm{xtan}\left(\alpha \right)-\left(\frac{{\mathrm{ax}}^2}{2{\mathrm{u}}^2{\cos }^2\left(\alpha \right)}\right)$and the tangent is parallel to $y=x$ has an ordinate $\frac{u^2}{4a}$,

$\Rightarrow \frac{dy}{dx}=\tan \left(\alpha \right)–\frac{2ax}{2u^2{\cos }^2\left(\alpha \right)}$

Since the tangent is parallel to $y=x$ then, $\frac{dy}{dx}=1$.

$\Rightarrow 1=\frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}-\frac{ax}{u^2{\cos }^2\left(\alpha \right)}\left[\because \frac{dy}{dx}=1\right]$

Multiplying both side by ${\cos }^2\left(\alpha \right)$

$$\begin{gathered} \Rightarrow {\cos }^2\left(\alpha \right)=\frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}·{\cos }^2\left(\alpha \right)-\frac{ax}{u^2{\cos }^2\left(\alpha \right)}·{\cos }^2\left(\alpha \right) \\ \Rightarrow {\cos }^2\left(\alpha \right)=\sin \left(\alpha \right)\cos \left(\alpha \right)-\frac{ax}{u^2} \\ \Rightarrow \frac{ax}{u^2}=\cos \left(\alpha \right)\left(\sin \left(\alpha \right)-\cos \left(\alpha \right)\right) \\ \Rightarrow x=\frac{u^2}{a}\cos \left(\alpha \right)\left(\sin \left(\alpha \right)-\cos \left(\alpha \right)\right) \end{gathered}$$

Step 2. Find the value of $y$

$\mathrm{y}=\mathrm{xtan}\left(\alpha \right)-\left(\frac{{\mathrm{ax}}^2}{2{\mathrm{u}}^2{\cos }^2\left(\alpha \right)}\right)$

Substituting the value of $x$

$$\begin{gathered} \therefore y=\frac{u^2}{a}\cos \left(\alpha \right)\left(\sin \left(\alpha \right)-\cos \left(\alpha \right)\right)\tan \left(\alpha \right)-\frac{a}{2u^2{\cos }^2\left(\alpha \right)}{\left[\frac{u^2}{a}\cos \left(\alpha \right)\left(\sin \left(\alpha \right)-\cos \left(\alpha \right)\right)\right]}^2 \\ =\frac{u^2}{a}\sin \left(\alpha \right)\left(\sin \left(\alpha \right)-\cos \left(\alpha \right)\right)-\frac{a}{2u^2{\cos }^2\left(\alpha \right)}·\frac{u^4}{a^2}{\cos }^2\left(\alpha \right)\left({\sin }^2\left(\alpha \right)+{\cos }^2\left(\alpha \right)-2\sin \left(\alpha \right)\cos \left(\alpha \right)\right)\left[\because \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\right] \\ =\frac{u^2}{a}\sin \left(\alpha \right)\left(\sin \left(\alpha \right)-\cos \left(\alpha \right)\right)-\frac{1}{2}·\frac{u^2}{a}\left(1-2\sin \left(\alpha \right)\cos \left(\alpha \right)\right)\left[\because {\sin }^2\left(\alpha \right)+{\cos }^2\left(\alpha \right)=1\right] \\ =\frac{u^2}{a}\left({\sin }^2\left(\alpha \right)-\sin \left(\alpha \right)\cos \left(\alpha \right)-\frac{1}{2}+\sin \left(\alpha \right)\cos \left(\alpha \right)\right) \\ =\frac{u^2}{a}\left({\sin }^2\left(\alpha \right)-\frac{1}{2}\right) \end{gathered}$$

Step 3. Find the angle $\alpha$

So we have $y=\frac{u^2}{a}\left({\sin }^2\left(\alpha \right)-\frac{1}{2}\right)$

Given ordinate $\frac{u^2}{4a}$

$\therefore y=\frac{u^2}{4a}$

$$\begin{gathered} \Rightarrow \frac{u^2}{4a}=\frac{u^2}{a}\left({\sin }^2\left(\alpha \right)-\frac{1}{2}\right) \\ \therefore \frac{1}{4}={\sin }^2\left(\alpha \right)-\frac{1}{2} \\ \Rightarrow \sin \left(\alpha \right)=\frac{\sqrt{3}}{2} \\ \Rightarrow \alpha =\frac{\pi }{3} \end{gathered}$$

Hence, option $\left(\mathrm{C}\right)$ is correct.