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

Hint: 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.