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The length of the perpendicular from the...

The length of the perpendicular from the origin, on the normal to the curve, `x^2 + 3xy - 10 y^2 = 0` at the point (2,1) is :

A

`3 sqrt5`

B

`3`

C

`sqrt5`

D

`2/sqrt5`

Text Solution

Verified by Experts

The correct Answer is:
C
`x^2 + 3xy -10 y^2 =0`
`m_N` = slope of normal drawn to curve at (2, 1) is -2
`L : 2x + y =5`
Perpendicular distance of L from (0, 0)
`=(|0+0-5|)/sqrt5 = 5/sqrt5`
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