If the tangent to the curve x^{2/3} + y^{2/3} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ ...$(i)$Let a point on the curve be $P(a \cos^3 	heta, a \sin^3 	heta)$.Differentiating the cur

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(C) Given the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ ...$(i)$Let a point on the curve be $P(a \cos^3 	heta, a \sin^3 	heta)$.Differentiating the curve $(i)$ with respect to $x$,we get:$\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\left(\frac{y}{x}ight)^{1/3}$.Substituting the point $P$,the slope of the tangent $m$ is:$m = -\left(\frac{a \sin^3 	heta}{a \cos^3 	heta}ight)^{1/3} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$.The equation of the tangent at point $P$ is:$y - a \sin^3 	heta = -	an 	heta (x - a \cos^3 	heta)$.$y \cos 	heta - a \sin^3 	heta \cos 	heta = -x \sin 	heta + a \cos^3 	heta \sin 	heta$.$x \sin 	heta + y \cos 	heta = a \sin 	heta \cos 	heta (\cos^2 	heta + \sin^2 	heta) = a \sin 	heta \cos 	heta$.For point $A$ (where $y=0$),$x \sin 	heta = a \sin 	heta \cos 	heta \Rightarrow x = a \cos 	heta$. So $A = (a \cos 	heta, 0)$.For point $B$ (where $x=0$),$y \cos 	heta = a \sin 	heta \cos 	heta \Rightarrow y = a \sin 	heta$. So $B = (0, a \sin 	heta)$.The distance $AB = \sqrt{(a \cos 	heta - 0)^2 + (0 - a \sin 	heta)^2} = \sqrt{a^2 \cos^2 	heta + a^2 \sin^2 	heta} = \sqrt{a^2} = a$.Thus,$AB = a$.

If the tangent to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ meets the $X$-axis at $A$ and $Y$-axis at $B$,then $AB =$

Similar Questions

Find the points on the curve $\frac{x^2}{4} + \frac{y^2}{25} = 1$ at which the tangents are parallel to the $x$-axis.

If $\theta$ is an angle between the curves $x^2+4y=0$ and $xy=2$,then $\tan \theta=$

$p_1$ and $p_2$ are the perpendicular distances from the origin to the tangent and normal drawn at any point on the curve $x^{2/3} + y^{2/3} = a^{2/3}$ respectively. If $k_1 p_1^2 + k_2 p_2^2 = a^2$,then $k_1 + k_2 =$

The equation of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at the point $(x_1, y_1)$ is:

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $P(-2,0)$ and cuts the $Y$-axis at a point $Q$,where the gradient is $3$. Then the values of $a, b, c$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module