0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# Find the equation of the lines joining the origin to the points of intersection of the straight line y=3x+2 with the curve x2+2xy+3y2+4x+8y−11=0

A

7x22xy+y2=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

7x22xyy2=0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

7x2+2xy+y2=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

7x2+2xy+y2=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 7x2−2xy−y2=0 We will use method of homogenization to solve this. The combination of pairs straight lines is given by Ax2+2hxy+by2+2gx(lx+my−n)+2fy(lx+my−n)+c(lx+my−n)2=0 the term equivalent to lx+my−n is y−3x2. We can take (3x−y−2) also. ⇒x2+2xy+3y2+4x(y−3x2)+8x(y−3x2)-11x(y−3x2)2 ⇒x2+2xy+3y2+2xy−6x2+4y2−12xy−114(y2+9x2−6xy)=0 ⇒−5x2+7y2−8xy−11y2−99x2+66xy4=0 ⇒−20x2+28y2−32xy−11y2−99x2+66xy=0 ⇒−119x2+17y2+34xy=0 ⇒7x2−2xy−y2=0

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
MATHEMATICS
Watch in App
Join BYJU'S Learning Program