Question

If the normal to the curve \[{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}}\] makes an angle \[\phi \] with the \[x - axis\] , then its equation is :
A.\[{\text{x sin }}\phi {\text{ + y cos }}\phi {\text{ }} = {\text{ }}a\]
B.\[y{\text{ cos }}\phi {\text{ }} - {\text{ }}x{\text{ sin }}\phi {\text{ }} = {\text{ }}a{\text{ cos }}2\phi \]
C.\[{\text{x cos }}\phi {\text{ + y sin }}\phi {\text{ }} = {\text{ }}a{\text{ sin }}2\phi \]
D.none of these

Answer 1

Hint: We have to find the equation of the curve normal to \[{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}}\] which makes an angle \[\phi \] with the \[x - axis\] . We solve this question using the concept of equation of line and the slope of the line . We should also have the knowledge about the concept of normal to the curve and the tangent of the curve.

Complete step-by-step answer:
Given : \[{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}} - - - - (1)\]
The slope of the curve is given as :
\[slope{\text{ }} = {\text{ }}\dfrac{{dy}}{{dx}}\]
So , we have to differentiate \[(1)\] with respect to ‘ \[x\] ’ .
Differentiating y with respect to ‘ \[x\] ’ , we get
Using (Derivative of \[{x^n} = n \times {x^{n - 1}}\] )
(Derivative of \[constant{\text{ }} = {\text{ }}0\] )
\[\dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} + \dfrac{2}{3}{y^{\dfrac{{ - 1}}{3}}} \times \dfrac{{dy}}{{dx}} = 0\]
\[\dfrac{{dy}}{{dx}} = {\left[ {\dfrac{{ - y}}{x}} \right]^{\dfrac{1}{3}}}\]
Let \[P\left( {h{\text{ }},{\text{ }}k} \right)\] be any point on the curve \[(1)\]
Slope of the normal at \[P\] is \[m = {\left( {\dfrac{h}{k}} \right)^{\dfrac{1}{3}}}\]
It is given that, the slope of the normal \[ = \tan \phi \]
\[{\left( {\dfrac{h}{k}} \right)^{\dfrac{1}{3}}} = \tan \phi \]
As , \[{\text{tan x}} = \dfrac{{\sin x}}{{\cos x}}\]
\[{\left( {\dfrac{h}{k}} \right)^{\dfrac{1}{3}}} = \dfrac{{\sin \phi }}{{\cos \phi }}\]
\[\dfrac{{{h^{\dfrac{1}{3}}}}}{{\sin \phi }} = \dfrac{{{k^{\dfrac{1}{3}}}}}{{\cos \phi }}\]
Let ,
\[\dfrac{{{h^{\dfrac{1}{3}}}}}{{\sin \phi }} = \dfrac{{{k^{\dfrac{1}{3}}}}}{{\cos \phi }} = {a^{\dfrac{1}{3}}}\]
So ,
\[\dfrac{{{h^{\dfrac{1}{3}}}}}{{\sin \phi }} = {a^{\dfrac{1}{3}}}\] and \[\dfrac{{{k^{\dfrac{1}{3}}}}}{{\cos \phi }} = {a^{\dfrac{1}{3}}}\]
Cubing both sides , we get
\[h = a{\sin ^3}\phi \] and \[k = a{\cos ^3}\phi \]
The equation of the normal is given as :
\[\left( {y - k} \right) = m\left( {x - h} \right)\]
Where m is the slope of the equation
Substituting the values in the equation of line , we get
\[y - a{\cos ^3}\phi = \tan \phi \times (x - a{\sin ^3}\phi )\]
As , \[{\text{tan x}} = \dfrac{{\sin x}}{{\cos x}}\]
\[y - a{\cos ^3}\phi = \dfrac{{\sin \phi }}{{\cos \phi }} \times (x - a{\sin ^3}\phi )\]
On simplifying , we get
\[y\cos \phi - a{\cos ^4}\phi = \sin \phi \times x - a{\sin ^4}\phi \]
\[x\sin \phi - y\cos \phi = a{\sin ^4}\phi - a{\cos ^4}\phi \]
Taking \[a\] common , we get
\[x\sin \phi - y\cos \phi = a[{\sin ^4}\phi - {\cos ^4}\phi ]\]
We , know that \[{a^2} - {b^2} = (a + b)(a - b)\]
So , we can express it as
\[x\sin \phi - y\cos \phi = a[{\sin ^2}\phi - {\cos ^2}\phi ][{\sin ^2}\phi + {\cos ^2}\phi ]\]
We know that , \[[{\sin ^2}\phi + {\cos ^2}\phi ] = 1\]
So ,
\[x\sin \phi - y\cos \phi = a[{\sin ^2}\phi - {\cos ^2}\phi ]\]
Also , we know that \[\cos 2x = {\cos ^2}x - {\sin ^2}x\]
\[x\sin \phi - y\cos \phi = - a\cos 2\phi \]
Taking \[(-1)\] common , we get
\[y\cos \phi - x\sin \phi = a\cos 2\phi \]
Thus , the equation of the curve is \[y\cos \phi - x\sin \phi = a\cos 2\phi \] .
Hence , the correct option is \[(2)\] .
So, the correct answer is “Option 2”.

Note: If \[\dfrac{{dy}}{{dx}}\] does not exist at the point \[\left( {{x_0},{\text{ }}{y_0}} \right)\] , then the tangent at this point is parallel to the \[y - axis\] and its equation is \[x = {x_0}\] .
If tangent to a curve \[y{\text{ }} = {\text{ }}f\left( x \right)\] at \[x = {x_0}\] is parallel to \[x{\text{ }} - {\text{ }}axis\] , then \[\dfrac{{dy}}{{dx}}\] at \[(x = {x_0}) = {\text{ }}0\] .
Slope of the equation is also given as \[m = \tan {\text{ }}x\] , where \[x\] is the angle which the line makes with the positive \[x{\text{ }} - {\text{ }}axis\] .