Please explain the following.....how it's working.
Suppose a curve 3x2-y2-2x+4y=0 has all chords subtending a right angle at the origin.
Let equation of chord be lx+my=1
To make the equation of the curve HOMOGENEOUS
3x2-y2-2(x-2y)(lx+my)=0
or,(3-2l)x2+(4m-1)y2+(4l-2m)xy=0......(1)
First question , what is homogeneous eqn. and what is the need to make the eqn. of the curve homogeneous ?
2nd. It is said that eqn.(1) represents two perpendicular lines....basics?
3rd. hence show that all the chords subtending a rt. angle at the origin are concurrent.
Please clear the basics on which this question stands ?
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1 Answers
1357or,(3-2l)x2+(4m-1)y2+(4l-2m)xy=0......(1)
3rd. hence show that all the chords subtending a rt. angle at the origin are concurrent.
Homogeneous is related with the degree of the terms
here making it homogeneous, you can directly solve for (y/x) which is a quadratic equation
so it is of the form (y/x)=k and y/x=c
so it gives two straight lines passing through origin
By homogenizing it means it passes through the intersection of the chord and the curve (see this yourself)
SO equation 1 represents two straight lines passing through origin and through the intersection of chord and the curve.
Let the angle between the two lines be α
From the given condition, α=90o
