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 $x{\text{ }} - {\text{ }}axis$ .