The length of the perpendicular from the origin, on the normal to the curve, $x^2+2 x y-3 y^2=0$ at the point $(2,2)$ is
Select the correct option:
A
$2 \sqrt{2}$
B
$\sqrt{2}$
C
$4 \sqrt{2}$
D
2
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Given curve x² + 3xy − xy − 3y² = 0
⇒ (x − y)(x + 3y) = 0 ⇒ x and y = 3
Normal pass through (2, 2) and is perpendicular to line x − y = 0
Let normal is x + y + λ = 4
Perpendicular distance = |λ| / √2 = 2√2
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