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If x^2+2h x y+y^2=0 represents the equat...

If `x^2+2h x y+y^2=0` represents the equation of the straight lines through the origin which make an angle `alpha` with the straight line `y+x=0` then, (a) `s e c2alpha=h` (b) `cosalpha` `=sqrt(((1+h))/((2h)))` (c) `2sinalpha` `=sqrt(((1+h))/h)` (d) `cotalpha` `=sqrt(((1+h))/((h-1)))`

Text Solution

Verified by Experts

The correct Answer is:
1,2,4
Let the equation of the lines given by `x^(2)+2hxy+y^(2)=0` be `y=m_(1)xandy=m_(2)x`. Since these make an angle `alpha` with `y +x=0` whose slope is -1, we have
`(m_(1)+1)/(1-m_(1))=tanalpha=(-1-m_(2))/(1-m_(2))`
or `m_(1)+m_(2)=((tanalpha-1)^(2)+(tanalpha+1)^(2))/(tan^(2)alpha-1)`
`=(-sec^(2)alphaxxcos^(2)alpha)/(cos2alpha)`
`:.-2sec2alpha=-2h`
or `sec2alpha=h`
or `cos2alpha=(1)/(h)or2cos^(2)alpha-1=(1)/(h)`
or `cosalpha=sqrt((1+h)/(2h))andcotalpha=sqrt((h+1)/(h-1))`
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