Question

write the nature of line x^2-2xy-3y^2=0​

Answer 1

Answer:

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Step-by-step explanation:

Answer

x

2

+2xy−3y

2

=0

This is the equation of a pair of straight lines.

(x−y)(x+3y)=0

2x+2y+2xy

′

−6yy

′

=0

⇒

3y−x

x+y

=y

′

At (1,1)

the slope of the tangent is y

′

=1.

The slope of the normal is −1.

Hence,

The normal has the equation : x+y=2

We need to find its intersection with x+3y=0

On solving we get,

2−y+3y=0⇒2y=−2

y=−1,x=3

Hence, it meets it in the fourth quadrant.