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3.3 · Q1

Q.Find the slopes of the tangents and normal to the curves at the indicated points. i. y=x3−xy = x^3 - x at x=1x = 1.
ii. y=3x2−6xy = 3x^2 - 6x at x=2x = 2.
iii. y=x−1x−2, x≠2y = \dfrac{x-1}{x-2},\ x \neq 2 at x=10x = 10.
iv. x2/3+y2/3=2x^{2/3} + y^{2/3} = 2 at (1, 1).

Delhi CbseNCERTSubjective· 5mImportance★★★★★
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The tangent slope is dydx\frac{dy}{dx} at the point and the normal slope is −1(dy/dx)-\frac{1}{(dy/dx)}: (i) 2,−122,-\frac12;

(ii) 6,−166,-\frac16;

(iii) −164,64-\frac{1}{64},64;

(iv) −1,1-1,1.

Slope of tangent =mT=dydx∣point=m_T=\dfrac{dy}{dx}\Big|_{\text{point}}; slope of normal =mN=−1mT=m_N=-\dfrac{1}{m_T} (since tangent ⊥\perp normal).

  1. (i) y=x3−xy=x^3-x at x=1x=1:

dydx=3x2−1  ⇒  mT=3(1)2−1=2,mN=−12.\frac{dy}{dx}=3x^2-1\;\Rightarrow\;m_T=3(1)^2-1=2,\qquad m_N=-\frac{1}{2}.

  1. (ii) y=3x2−6xy=3x^2-6x at x=2x=2:

dydx=6x−6  ⇒  mT=6(2)−6=6,mN=−16.\frac{dy}{dx}=6x-6\;\Rightarrow\;m_T=6(2)-6=6,\qquad m_N=-\frac{1}{6}.

  1. (iii) y=x−1x−2y=\dfrac{x-1}{x-2} at x=10x=10 (quotient rule):

dydx=(x−2)(1)−(x−1)(1)(x−2)2=−1(x−2)2.\frac{dy}{dx}=\frac{(x-2)(1)-(x-1)(1)}{(x-2)^2}=\frac{-1}{(x-2)^2}.

mT=−1(10−2)2=−164,mN=−1mT=64.m_T=\frac{-1}{(10-2)^2}=\frac{-1}{64},\qquad m_N=-\frac{1}{m_T}=64.

  1. (iv) x2/3+y2/3=2x^{2/3}+y^{2/3}=2 at (1,1)(1,1) (implicit differentiation):

23x−1/3+23y−1/3dydx=0  ⇒  dydx=−x−1/3y−1/3=−(yx)1/3.\frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\frac{dy}{dx}=0\;\Rightarrow\;\frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}=-\left(\frac{y}{x}\right)^{1/3}.

mT=−(11)1/3=−1,mN=−1−1=1.m_T=-\left(\frac{1}{1}\right)^{1/3}=-1,\qquad m_N=-\frac{1}{-1}=1.

✓Final answer

  1. tangent 22, normal −12-\dfrac12;
  2. tangent 66, normal −16-\dfrac16;
  3. tangent −164-\dfrac{1}{64}, normal 6464;
  4. tangent −1-1, normal 11.

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