JEE Main 2020 — Application Of Derivatives Question with Solution
From: JEE Main 2020 (Online) 8th January Evening Slot
Question
The length of the perpendicular from the origin,
on the normal to the curve,
x2 + 2xy – 3y2 = 0 at the point (2,2) is
x2 + 2xy – 3y2 = 0 at the point (2,2) is
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
x2 + 2xy – 3y2 = 0
Differentiate the curve
2x + 2y + 2xy' – 6yy' = 0
x + y + xy' – 3yy' = 0
y'(x – 3y) = – (x + y)
y' =
Slope of normal = =
= = -1
Normal at (2, 2)
y – 2 = – 1 (x – 2)
y + x = 4
Perpendicular distance from (0,0)
= =
Differentiate the curve
2x + 2y + 2xy' – 6yy' = 0
x + y + xy' – 3yy' = 0
y'(x – 3y) = – (x + y)
y' =
Slope of normal = =
= = -1
Normal at (2, 2)
y – 2 = – 1 (x – 2)
y + x = 4
Perpendicular distance from (0,0)
= =
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