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Worked Examples · Example 20

Q.Find the equation of the tangent to the curve x2+3y−3=0x^2 + 3y - 3 = 0, which is parallel to the line y=4x−5y = 4x - 5.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
23% · 20/87 Questions
✓ Free question

A tangent parallel to y=4x−5y=4x-5 has slope 44; set the curve's slope equal to 44 to locate the point of contact, then write the line.

Parallel lines have equal slopes: mt=4m_t=4. Line: y−y0=m(x−x0)y-y_0=m(x-x_0).

  • mtm_t = slope of the tangent = dydx\dfrac{dy}{dx} at the point of contact.
  1. Rewrite the curve: x2+3y−3=0⇒y=1−x23x^2+3y-3=0\Rightarrow y=1-\dfrac{x^2}{3}.
  2. Differentiate:

dydx=−2x3.\frac{dy}{dx}=-\frac{2x}{3}.

  1. Tangent is parallel to y=4x−5y=4x-5 (slope 44), so

−2x3=4 ⇒ x=−6.-\frac{2x}{3}=4\ \Rightarrow\ x=-6.

  1. Find yy: y=1−(−6)23=1−363=1−12=−11y=1-\dfrac{(-6)^2}{3}=1-\dfrac{36}{3}=1-12=-11. Point of contact (−6,−11)(-6,-11).
  2. Tangent through (−6,−11)(-6,-11) with slope 44:

y−(−11)=4(x−(−6)) ⇒ y+11=4x+24 ⇒ y=4x+13.y-(-11)=4(x-(-6))\ \Rightarrow\ y+11=4x+24\ \Rightarrow\ y=4x+13.

✓Final answer

Tangent: y=4x+13y=4x+13, i.e. 4x−y+13=04x-y+13=0, at the point (−6,−11)(-6,-11).

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