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Worked Examples · Example 18

Q.Find the equation of the tangent and normal to the curve x2/3+y2/3=2x^{2/3} + y^{2/3} = 2 at (1, 1).

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
21% · 18/87 Questions
✓ Free question

Differentiate implicitly to get the slope at (1,1)(1,1), then write the tangent and normal lines through the point.

Implicit differentiation of x2/3+y2/3=2x^{2/3}+y^{2/3}=2; tangent slope m=dydxm=\dfrac{dy}{dx}, normal slope =−1m=-\dfrac1m; line through (x0,y0)(x_0,y_0): y−y0=m(x−x0)y-y_0=m(x-x_0).

  1. Differentiate x2/3+y2/3=2x^{2/3}+y^{2/3}=2 w.r.t. xx:

23x−1/3+23y−1/3dydx=0.\dfrac{2}{3}x^{-1/3}+\dfrac{2}{3}y^{-1/3}\dfrac{dy}{dx}=0.

  1. Solve for the slope:

dydx=−x−1/3y−1/3=−(yx)1/3.\dfrac{dy}{dx}=-\dfrac{x^{-1/3}}{y^{-1/3}}=-\left(\dfrac{y}{x}\right)^{1/3}.

  1. Evaluate at (1,1)(1,1):

m=−(11)1/3=−1.m=-\left(\dfrac{1}{1}\right)^{1/3}=-1.

  1. Tangent line through (1,1)(1,1):

y−1=−1(x−1) ⇒ y−1=−x+1 ⇒ x+y=2.y-1=-1(x-1)\ \Rightarrow\ y-1=-x+1\ \Rightarrow\ x+y=2.

  1. Normal slope =−1m=−1−1=1=-\dfrac{1}{m}=-\dfrac{1}{-1}=1. Normal line:

y−1=1(x−1) ⇒ y=x ⇒ x−y=0.y-1=1(x-1)\ \Rightarrow\ y=x\ \Rightarrow\ x-y=0.

✓Final answer

Tangent: x+y=2x+y=2; Normal: x−y=0x-y=0 (that is, y=xy=x).

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