Question

If the point on $\,\,y=x\tan (\alpha )-\left( \dfrac{a{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,,\,(a > 0)$ where the tangent is parallel to $\,y=x$ has an ordinate $\,\,\dfrac{{{u}^{2}}}{4a}$ then $\,\alpha$ is equal to:
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{3}$
D. $\dfrac{\pi }{2}$

Answer 1

Here, in the question $\,\,y=x\tan (\alpha )-\left( \dfrac{a{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)$ is given and also given that tangent is parallel to $\,y=x$ that means $\dfrac{dy}{dx}=1$ . We have to take the derivative of the above equation and equate the value of $\,\,x$ with ordinate which is given in the question that is $\,\,\dfrac{{{u}^{2}}}{4a}\,$ .

Complete step by step answer:
According to the given equation:
$\,\,y=x\tan (\alpha )-\left( \dfrac{a{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,---(1)$
According to the given condition that is tangent which is parallel to $\,y=x$ , $\dfrac{dy}{dx}=1----(2)$
Taking derivative on both sides on equation $(1)$
$\dfrac{dy}{dx}=\tan (\alpha )-\left( \dfrac{2ax}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,$
2 get cancelled after further simplifying we get:
$\dfrac{dy}{dx}=\tan (\alpha )-\left( \dfrac{ax}{{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,----(3)$
By using the trigonometry property, $\therefore \tan (\alpha )=\dfrac{\sin (\alpha )}{\cos (\alpha )}---(4)$
By substituting the value of equation $(4)$ in equation $(3)$ , we get:
$\dfrac{dy}{dx}=\dfrac{\sin (\alpha )}{\cos (\alpha )}-\left( \dfrac{ax}{{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,---(5)$
By substituting the value of equation $(2)$ in equation $(5)$ we get:
$1=\dfrac{\sin (\alpha )}{\cos (\alpha )}-\left( \dfrac{ax}{{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,---(6)$
Multiply ${{\cos }^{2}}(\alpha )$ on both side on equation $(6)$
${{\cos }^{2}}(\alpha )=\dfrac{\sin (\alpha )}{\cos (\alpha )}\times {{\cos }^{2}}(\alpha )-\left( \dfrac{ax}{{{u}^{2}}} \right)$
After further simplifying we get:
${{\cos }^{2}}(\alpha )=\sin (\alpha )\cos (\alpha )-\left( \dfrac{ax}{{{u}^{2}}} \right)\,$
Take $\sin (\alpha )\cos (\alpha )$ on left side and further simplifying we get:
$\left( \dfrac{ax}{{{u}^{2}}} \right)\,=\sin (\alpha )\cos (\alpha )-{{\cos }^{2}}(\alpha )$
In the above equation we have to find the value x by further simplifying we get:
$x=\dfrac{{{u}^{2}}}{a}\times (\sin (\alpha )\cos (\alpha )-{{\cos }^{2}}(\alpha ))---(7)$
By taking ${{\cos }^{2}}(\alpha )$ common on equation $(7)$ we get:
$x=\dfrac{{{u}^{2}}}{a}\times (\sin (\alpha )-\cos (\alpha ))\times \cos (\alpha )---(8)$
By substituting the value of equation $(8)$ in equation $(1)$
$y=\left( \dfrac{{{u}^{2}}}{a}\times (\sin (\alpha )-\cos (\alpha ))\times \cos (\alpha ) \right)\tan (\alpha )-\left( \dfrac{a}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,{{\left( \dfrac{{{u}^{2}}}{a}\times (\sin (\alpha )-\cos (\alpha ))\times \cos (\alpha ) \right)}^{2}}---(9)$
By substituting the equation $(4)$ in equation $(9)$
$y=\left( \dfrac{{{u}^{2}}}{a}\times (\sin (\alpha )-\cos (\alpha ))\times \cos (\alpha ) \right)\times \left( \dfrac{\sin (\alpha )}{\cos (\alpha )} \right)-\left( \dfrac{a}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,{{\left( \dfrac{{{u}^{2}}}{a}\times (\sin (\alpha )-\cos (\alpha ))\times \cos (\alpha ) \right)}^{2}}$
By simplifying further we get:
$y=\left( \dfrac{{{u}^{2}}}{a}\times (\sin (\alpha )-\cos (\alpha ))\times \sin (\alpha ) \right)-\left( \dfrac{a}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,\left( \dfrac{{{u}^{4}}}{{{a}^{2}}}\times {{(\sin (\alpha )-\cos (\alpha ))}^{2}}\times {{\cos }^{2}}(\alpha ) \right)---(10)$
By using the property of ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
${{(\sin (\alpha )-\cos (\alpha ))}^{2}}={{\sin }^{2}}(\alpha )-2\sin (\alpha )\cos (\alpha )+{{\cos }^{2}}(\alpha )--(11)$
By substituting the equation $(11)$ in equation $(10)$ and further simplifying we get:
$y=\left( \dfrac{{{u}^{2}}}{a}\times ({{\sin }^{2}}(\alpha )-\cos (\alpha )\sin (\alpha )) \right)-\left( \dfrac{a}{2{{u}^{2}}{{\cos }^{2}}(\alpha )} \right)\,\left( \dfrac{{{u}^{4}}}{{{a}^{2}}}\times ({{\sin }^{2}}(\alpha )-2\sin (\alpha )\cos (\alpha )+{{\cos }^{2}}(\alpha ))\times {{\cos }^{2}}(\alpha ) \right)$
Further simplifying we get:
$y=\left( \dfrac{{{u}^{2}}}{a}\times ({{\sin }^{2}}(\alpha )-\cos (\alpha )\sin (\alpha )) \right)-\left( \dfrac{1}{2} \right)\,\left( \dfrac{{{u}^{2}}}{a}\times ({{\sin }^{2}}(\alpha )-2\sin (\alpha )\cos (\alpha )+{{\cos }^{2}}(\alpha )) \right)$
By using trigonometry identity property, ${{\sin }^{2}}(\alpha )+{{\cos }^{2}}(\alpha )=1$
and also taking $\dfrac{{{u}^{2}}}{a}$ common on this above equation we get:
$y=\dfrac{{{u}^{2}}}{a}\times \left( ({{\sin }^{2}}(\alpha )-\cos (\alpha )\sin (\alpha ))-\left( \dfrac{1}{2} \right)\,\left( 1-2\sin (\alpha )\cos (\alpha ) \right) \right)$
By further simplifying we get:
$y=\dfrac{{{u}^{2}}}{a}\times \left( {{\sin }^{2}}(\alpha )-\dfrac{1}{2} \right)--(12)$
Ordinate is given in question that is $\,y=\dfrac{{{u}^{2}}}{4a}\,\,----(13)\,\,$
Substitute the value of equation $(13)\,$ in equation $(12)$
$\dfrac{{{u}^{2}}}{4a}\,=\dfrac{{{u}^{2}}}{a}\times \left( {{\sin }^{2}}(\alpha )-\dfrac{1}{2} \right)$
$\dfrac{{{u}^{2}}}{a}$ Get cancelled on both sides
After simplification we get:
$\dfrac{1}{4}\,={{\sin }^{2}}(\alpha )-\dfrac{1}{2}$
$\sin (\alpha )=\dfrac{\sqrt{3}}{2}$
After take $\sin (\alpha )$ on right hand side it becomes $\,{{\sin }^{-1}}(\alpha )$ we get the value of $\alpha$
$\alpha =\,\dfrac{\pi }{3}$

So, the correct answer is “Option C”.

Note: Remember that according to the conditions the tangent is parallel to $\,y=\,\,\,x$ that means its derivative should be one. The equation which is given in question in that value of $\,a>0$ . So the above solution which is given can be referred for similar types of problems.