Question

show that the straight lines
x=2y=8z; x=y= 2z; 42=7 y = 7z lie
on a circular cone of semi verticle angle
$(cos)^{- 1} \frac{11}{ \sqrt{126} }$
​

Answer 1

Answer:

for a general eq ax

2

+2hxy+by

2

+2gx+2fy+c=0 will represents pair of straight line if:

a,b,h



=0 and h

2

−ab>or=0

for 3x

2

+7xy+2y

2

+5x+5y+2=0 a=3,h=7/2,b=2

h

2

−ab=49/4−6=25/4>0 (represents straight line)

for 2x

2

−13xy−7y

2

+x+23y−6=0

a=2,h=−13/2,b=−7

h

2

−ab=169/4+14=225/4>0 (represents straight line)

Points of intersection is α=

ab−b

2

hf−bg

and β=

ab−h

2

hg−af

after substituting values of a,b,h,g,f we will get the points of intersection for the given lines.

acute angle θ=cos

−1

(a−b)

2

+4h

2

∣a+b∣

again, after putting the values of a,b,h will get the acute angle.