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3.3 · Q2

Q.Find the equations of the tangent and normal to the curves at the indicated points. i. y=x3−3x+5y = x^3 - 3x + 5 at the point (2, 7).
ii. x=at2x = at^2, y=2aty = 2at at t=2t = 2

Yanam CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

Slope of tangent =dydx=\left.\dfrac{dy}{dx}\right. at the point; the normal slope is its negative reciprocal. Then use point-slope form.

Tangent at (x0,y0)(x_0,y_0) with slope m=dydx∣(x0,y0)m=\left.\dfrac{dy}{dx}\right|_{(x_0,y_0)}: y−y0=m(x−x0)y-y_0=m(x-x_0).

Normal: y−y0=−1m(x−x0)y-y_0=-\dfrac{1}{m}(x-x_0).

Part (i): y=x3−3x+5y=x^3-3x+5 at (2,7)(2,7)

  1. Differentiate: dydx=3x2−3\dfrac{dy}{dx}=3x^2-3.
  2. Slope at x=2x=2: m=3(2)2−3=12−3=9m=3(2)^2-3=12-3=9.
  3. Tangent: y−7=9(x−2)⇒y=9x−18+7=9x−11y-7=9(x-2)\Rightarrow y=9x-18+7=9x-11, i.e. 9x−y−11=09x-y-11=0.
  4. Normal slope =−19=-\dfrac{1}{9}: y−7=−19(x−2)⇒9y−63=−(x−2)⇒x+9y−65=0y-7=-\dfrac{1}{9}(x-2)\Rightarrow 9y-63=-(x-2)\Rightarrow x+9y-65=0.

Part (ii): x=at2, y=2atx=at^2,\ y=2at at t=2t=2

  1. Point: x=a(2)2=4a, y=2a(2)=4ax=a(2)^2=4a,\ y=2a(2)=4a, so the point is (4a,4a)(4a,4a).
  2. dxdt=2at, dydt=2a⇒dydx=2a2at=1t\dfrac{dx}{dt}=2at,\ \dfrac{dy}{dt}=2a\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}.
  3. Slope at t=2t=2: m=12m=\dfrac{1}{2}.
  4. Tangent: y−4a=12(x−4a)⇒2y−8a=x−4a⇒x−2y+4a=0y-4a=\dfrac{1}{2}(x-4a)\Rightarrow 2y-8a=x-4a\Rightarrow x-2y+4a=0.
  5. Normal slope =−2=-2: y−4a=−2(x−4a)⇒y−4a=−2x+8a⇒2x+y−12a=0y-4a=-2(x-4a)\Rightarrow y-4a=-2x+8a\Rightarrow 2x+y-12a=0.
✓Final answer

(i) Tangent 9x−y−11=09x-y-11=0; Normal x+9y−65=0x+9y-65=0.

(ii) Tangent x−2y+4a=0x-2y+4a=0; Normal 2x+y−12a=02x+y-12a=0.

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