CBSE Class 12 Sample Paper for 2021 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Find the equation of the normal to the curve y = π₯ + 1/x, Β  x > 0 perpendicular to the line 3π₯ β 4π¦ = 7.

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### Transcript

Question 22 Find the equation of the normal to the curve y = π₯ + 1/π₯, x > 0 perpendicular to the line 3π₯ β 4π¦ = 7 Finding Slope y = x + 1/π₯ Differentiating both sides ππ¦/ππ₯ = 1 β 1/π₯^2 Now, Slope of Normal = (βπ)/(π β π/π^π ) Given that Normal is perpendicular to 3x β 4y = 7 So, Slope of Normal Γ Slope of Line = β1 (β1)/(1 β 1/π₯^2 ) Γ 3/4 = β1 3/4 = 1 β 1/π₯^2 1 β 1/π₯^2 = 3/4 1 β 3/4 = 1/π₯^2 1/4 = 1/π₯^2 x2 = 4 x = Β± 2 Since x > 0 β΄ x = 2 Finding y when x = 2 y = x + 1/π₯ y = 2 + 1/2 y = π/π Now, Slope of Normal is (βπ)/π and it passes through point (2, π/π) So, equation of Normal is (y β y1) = m (x β x1) y β 5/2 = β 4/3 (x β 2) y β 5/2 = β 4/3x + 8/3 Multiplying both sides by 6 6y β 6 Γ 5/2 = β6 Γ 4/3x + 6 Γ 8/3 6y β 15 = β8x + 16 8x + 6y = 31