Check sibling questions

The equation of normal to the curve 3x 2 – y 2 = 8 which is parallel to the line x + 3y = 8 is

(A) 3x – y = 8Β  Β  Β  Β  Β  Β  Β  Β (B) 3x + y + 8 = 0

(C) x + 3y Β± 8 = 0Β  Β  Β  Β  Β (D) x + 3y = 0

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Transcript

Question 9 The equation of normal to the curve 3x2 – y2 = 8 which is parallel to the line x + 3y = 8 is (A) 3x – y = 8 (B) 3x + y + 8 = 0 (C) x + 3y Β± 8 = 0 (D) x + 3y = 0 Since, the normal to the curve is parallel to the line π‘₯+3𝑦=8 ∴ Slope of normal = Slope of line So, finding slope of normal and slope of line Finding slope of normal "3" π‘₯"2 –" 𝑦"2 = 8" Differentiating w.r.t. x 6π‘₯ βˆ’ 2𝑦 𝑑𝑦/𝑑π‘₯ = 0 2𝑦 𝑑𝑦/𝑑π‘₯ = 6π‘₯ 𝑑𝑦/𝑑π‘₯ =(6π‘₯ )/2𝑦 𝑑𝑦/𝑑π‘₯ =3π‘₯/𝑦 Slope of normal =(βˆ’1)/(𝑑𝑦/𝑑π‘₯) =(βˆ’1)/(3π‘₯/𝑦) =(βˆ’π’š)/πŸ‘π’™ Finding slope of line π‘₯+3𝑦=8 Differentiating w.r.t. x 1+3 𝑑𝑦/𝑑π‘₯ = 0 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/3 Slope of line =𝑑𝑦/𝑑π‘₯ =(βˆ’πŸ)/πŸ‘ ∴ Equating (1) & (2) (βˆ’π’š)/πŸ‘π’™ = (βˆ’πŸ)/πŸ‘ βˆ’3𝑦=βˆ’3π‘₯ π’š=𝒙 Now, to find equation of normal, we need a point So, Putting π’š=𝒙 in the curve 3𝑦^2βˆ’π‘₯^2=8 3π‘₯^2βˆ’π‘₯^2=8 2π‘₯^2 = 8 π‘₯^2 = 8/2 π‘₯^2 = 4 𝒙=±𝟐 For y-coordinates, putting value of π‘₯ in y=π‘₯ Finding equation of normal Equation of line at (π‘₯_1, 𝑦_1) & having slope m is (π‘¦βˆ’π‘¦_1 ) = m (π‘₯βˆ’π‘₯_1 ) For 𝒙 = 2 𝑦=π‘₯ 𝑦=2 So, the point is (2, 2) For 𝒙 = βˆ’2 𝑦=π‘₯ 𝑦=βˆ’2 So, the point is (βˆ’2, βˆ’2) Equation of normal at (2, 2) & Slope (βˆ’πŸ)/πŸ‘ (π‘¦βˆ’2) = (βˆ’1)/3 (π‘₯βˆ’2) 3 (π‘¦βˆ’2) = βˆ’1 (π‘₯βˆ’2) 3π‘¦βˆ’6=βˆ’π‘₯+2 3𝑦+π‘₯=6+2 3𝑦+π‘₯=8 πŸ‘π’š+π’™βˆ’πŸ–=𝟎 Equation of normal at (βˆ’2, βˆ’2) & Slope (βˆ’πŸ)/πŸ‘ (𝑦+2) = (βˆ’1)/3 (π‘₯+2) 3 (𝑦+2) = βˆ’1 (π‘₯+2) 3𝑦+6=βˆ’π‘₯βˆ’2 3𝑦+π‘₯=βˆ’6βˆ’2 3𝑦+π‘₯=βˆ’8 πŸ‘π’š+𝒙+πŸ–=𝟎 Hence, the required equation of normal is 3y + 𝒙 Β± 8 = 0 So, the correct answer is (C)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.