Question

(CBSE 20
Find the equations of the tangent and normal to the curve x2/3 + y213 = 2 at the point (1, 1).
INCE
Find the equation of the tangent and normal to the mm2​

Answer 1

EXPLANATION.

$\sf : \implies \: = equation \: of \: the \: \: tangent \: and \: the \: normal \: to \: the \: curve \\ \\ \sf : \implies \: {x}^{ \dfrac{2}{3} } + y {}^{ \dfrac{2}{3} } = 2 \: \: at \: \: point \: (1,1)$

$\sf : \implies \: differentiate \: w.r.t \: \: x \: \: we \: \: get, \\ \\ \sf : \implies \: \frac{2}{3}x {}^{ (\dfrac{ - 1}{3} )} + \frac{2}{3}y {}^{ (\dfrac{ - 1}{3} )} \frac{dy}{dx} = 0 \\ \\ \sf : \implies \: \frac{dy}{dx} = \frac{ - x {}^{ \dfrac{ - 1}{3} } }{y {}^{ \dfrac{ - 1}{3} } } = \frac{ - y {}^{ \dfrac{1}{3} } }{x {}^{ \dfrac{1}{3} } }$

$\sf : \implies \: put \: the \: value \: of \: x \: \: and \: \: y \: \\ \\ \sf : \implies \: \frac{dy}{dx} = - 1 \\ \\ \sf : \implies \: slope \: of \: the \: equation \: = - 1 \\ \\ \sf : \implies \: equation \: of \: tangent \\ \\ \sf : \implies \: (y - y_{1}) = m(x - x_{1})$

$\sf : \implies \: (y - 1) = - 1(x - 1) \\ \\ \sf : \implies \: y - 1 = - x + 1 \\ \\ \sf : \implies \: y + x - 2 = 0$

$\sf : \implies \: equation \: of \: normal \\ \\ \sf : \implies \: (y - y_{1}) = \frac{ - 1}{m} (x - x_{1}) \\ \\ \sf : \implies \: (y - 1) = 1(x - 1) \\ \\ \sf : \implies \: y - 1 = x - 1 \\ \\ \sf : \implies \: y - x = 0$

Answer 2

Given:

$\rm \bullet {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = 2$

Find:

• Eq. of Tangent and normal to the curve at the point (1,1)

Solution:

The curve is given as

$:\to \rm {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = 2$

$\qquad$ Differentiate above writ x

$\rm \dashrightarrow \dfrac{2}{3} {x}^{ \frac{ - 1}{3} } + \dfrac{2}{3} {y}^{ \frac{ - 1}{3} } \dfrac{dy}{dx} = 0$

$\sf \implies \dfrac{2}{3} {y}^{ \frac{ - 1}{3} } \dfrac{dy}{dx} = \dfrac{ - 2}{3} {x}^{ \frac{ - 1}{3} }$

$\sf \implies \dfrac{dy}{dx} = \dfrac{ - 2}{3} {x}^{ \frac{ - 1}{3} } \times \dfrac{3}{2} {y}^{ \frac{1}{3} }$

$\sf \implies \dfrac{dy}{dx} = \dfrac{ - 2}{3} \times {x}^{ \frac{ - 1}{3} } \times \dfrac{3}{2} {y}^{ \frac{1}{3} }$

$\sf \implies \dfrac{dy}{dx} = - {x}^{ \frac{ - 1}{3} } \times{y}^{ \frac{1}{3} }$

$\sf \implies \dfrac{dy}{dx} = - \dfrac{{y}^{ \frac{1}{3} }}{ {x}^{ \frac{1}{3} } } \qquad \big\lgroup{ \because {a}^{ - 2} = \dfrac{1}{ {a}^{2} } } \big\rgroup$

$\sf \implies \dfrac{dy}{dx} = - {\bigg(\dfrac{y}{x} \bigg)}^{ \frac{1}{3} } \qquad \big\lgroup{ \because \dfrac{{a}^{2}}{ {b}^{2} }= {\bigg(\dfrac{a}{ {b}} } \bigg)}^{2} \big\rgroup$

$\sf \implies \dfrac{dy}{dx} = - {\bigg(\dfrac{y}{x} \bigg)}^{ \frac{1}{3} }$

$\to\sf Slope \: of \: tangent \: to \: curve= - {\bigg(\dfrac{y}{x} \bigg)}^{ \frac{1}{3} } \: at \: point(1,1)$

Slope = - 1

Hence, Equation of Tangent at point (1,1) with slope - 1 is

$\rm\implies y - 1 = - 1(x - 1)$

$\rm\big \lgroup{ \therefore using \: y - y_{1} = m(x - x_{1}) } \big \rgroup$

$\rm\implies y - 1 = - 1(x - 1)$

$\rm\implies y - 1 = - x + 1$

$\rm\implies y - 1 - 1= - x$

$\rm\implies y - 2= - x$

$\rm\implies x + y - 2=$

$\rm\implies x + y - 2= 0$

$\rule{289}{1.6}$

we, know that

$\rm \to Slope \: of \: Normal = \dfrac{ - 1}{Slope \: of \: tangent} = 1$

So,

Eq. of Normal at (1,1) with slope 1 is

$\rm :\dashrightarrow (y - 1) = (x - 1)$

$\rm :\dashrightarrow y - 1 = x - 1$

$\rm :\dashrightarrow y = x$

$\rm :\dashrightarrow y - x = 0$

$\rule{289}{2}$

$\rm \therefore Equation \: of \: tangent \: is \: x + y - 2 = 0$

$\rm and \: Equation \: of \: normal \: is \: y - x= 0$