Question

Let $x^{2/3}+y^{2/3}=a^{2/3},(a>0)$ be the equation of curve then which of the following is/are correct?

• A. Equation of tangent at $(x_1,y_1)$ on the curve is $\frac{x}{x_1^{1/3}}+\frac{y}{y_1^{1/3}}=a^{2/3}$
• B. Length of portion of tangent intercepted between the coordinate axes is constant.
• C. If the curve $x^{2/3}+y^{2/3}=a^{2/3}$ touches the curve $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ then $c+d=a$
• D. Equation of normal at $(x_1,y_1)$ on the curve is $x.x_1^{1/3}-y.y_1^{1/3}=x_1^{4/3}+y_1^{4/3}$
  

Answer 1

Answer: (A), (B), (C)

The correct options are
A Equation of tangent at $(x_1,y_1)$ on the curve is $\frac{x}{x_1^{1/3}}+\frac{y}{y_1^{1/3}}=a^{2/3}$
B Length of portion of tangent intercepted between the coordinate axes is constant.
C If the curve $x^{2/3}+y^{2/3}=a^{2/3}$ touches the curve $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ then $c+d=a$

$(x,y)=(a{\cos }^3\theta ,a{\sin }^3\theta )$
$\frac{dy}{dx}=\frac{-\sin \theta }{\cos \theta }$
Equation of tangent $\to (y-a{\sin }^3\theta )=\frac{-\sin \theta }{\cos \theta }(x-a{\cos }^3\theta )$
$\to \frac{x}{\cos \theta }+\frac{y}{\sin \theta }=a$
$\to \frac{x}{x_1^{1/3}}+\frac{y}{y_1^{1/3}}=a^{2/3}$


Length of the portion of the tangent between the coodinate axes : $AB=\sqrt{a^2{\cos }^2\theta +a^2{\sin }^2\theta }=a$

Tangent at $\left(x_1{y}_1\right)to\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ is
$\frac{x{x}_1}{c^2}+\frac{y{y}_1}{d^2}=1$
which is identical to $\frac{x}{x_1^{1/3}}+\frac{x}{y_1^{1/3}}=a^{2/3}$
So $\frac{x_1}{c^2.}\times x_1^{1/3}=\frac{y_1}{c{l}^2}\times y_1^{1/3}=\frac{1}{a^{2/3}}$
$x_1^{4/3}=\frac{c^2}{a^{2/3}},y_1^{4/3}=\frac{d^2}{a^{2/3}}$
$x_1^{2/3}=\frac{c}{a^{1/3}},y_1^{2/3}=\frac{d}{a^{2/3}}$

which lies on $x^{2/3}+y^{2/3}=a^{2/3}$
$\frac{c}{a^{1/3}}+\frac{d}{a^{1/3}}=a^{2/3}$
$c+d=a$

Equation of normal
$y^{\prime }=\frac{y^{1/3}}{x^{1/3}}$
So normal is
$(y-y_1)=\frac{x_1^{1/3}}{y^{1/3}}(x-x_1)$

$x{x}_1^{1/3}-y{y}_1^{1/3}=x_1^{4/3}-y_1^{4/3}$