CBSE Class 12 Sample Paper for 2021 Boards
You are here
CBSE Class 12 Sample Paper for 2021 Boards
Last updated at Dec. 16, 2024 by Teachoo
Something went wrong!
The
video
couldn't load due to a technical hiccup.
But don't worry — our team is already on it, and we're working hard to get it back up ASAP.
Thanks for bearing with us!
Transcript
Question 22 Find the equation of the normal to the curve y = 𝑥 + 1/𝑥, x > 0 perpendicular to the line 3𝑥 − 4𝑦 = 7 Finding Slope y = x + 1/𝑥 Differentiating both sides 𝑑𝑦/𝑑𝑥 = 1 − 1/𝑥^2 Now, Slope of Normal = (−𝟏)/(𝟏 − 𝟏/𝒙^𝟐 ) Given that Normal is perpendicular to 3x − 4y = 7 So, Slope of Normal × Slope of Line = −1 (−1)/(1 − 1/𝑥^2 ) × 3/4 = −1 3/4 = 1 − 1/𝑥^2 1 − 1/𝑥^2 = 3/4 1 − 3/4 = 1/𝑥^2 1/4 = 1/𝑥^2 x2 = 4 x = ± 2 Since x > 0 ∴ x = 2 Finding y when x = 2 y = x + 1/𝑥 y = 2 + 1/2 y = 𝟓/𝟐 Now, Slope of Normal is (−𝟒)/𝟑 and it passes through point (2, 𝟓/𝟐) So, equation of Normal is (y − y1) = m (x − x1) y − 5/2 = − 4/3 (x − 2) y − 5/2 = − 4/3x + 8/3 Multiplying both sides by 6 6y − 6 × 5/2 = −6 × 4/3x + 6 × 8/3 6y − 15 = −8x + 16 8x + 6y = 31
