CBSE Class 12 Sample Paper for 2021 Boards
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CBSE Class 12 Sample Paper for 2021 Boards
Last updated at August 10, 2026 by Teachoo
Transcript
Question 22 Find the equation of the normal to the curve y = š„ + 1/š„, x > 0 perpendicular to the line 3š„ ā 4š¦ = 7 Finding Slope y = x + 1/š„ Differentiating both sides šš¦/šš„ = 1 ā 1/š„^2 Now, Slope of Normal = (āš)/(š ā š/š^š ) Given that Normal is perpendicular to 3x ā 4y = 7 So, Slope of Normal Ć Slope of Line = ā1 (ā1)/(1 ā 1/š„^2 ) Ć 3/4 = ā1 3/4 = 1 ā 1/š„^2 1 ā 1/š„^2 = 3/4 1 ā 3/4 = 1/š„^2 1/4 = 1/š„^2 x2 = 4 x = Ā± 2 Since x > 0 ā“ x = 2 Finding y when x = 2 y = x + 1/š„ y = 2 + 1/2 y = š/š Now, Slope of Normal is (āš)/š and it passes through point (2, š/š) So, equation of Normal is (y ā y1) = m (x ā x1) y ā 5/2 = ā 4/3 (x ā 2) y ā 5/2 = ā 4/3x + 8/3 Multiplying both sides by 6 6y ā 6 Ć 5/2 = ā6 Ć 4/3x + 6 Ć 8/3 6y ā 15 = ā8x + 16 8x + 6y = 31
