AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The given equation of line is,
3x+4y−5=0

3x+4y

5

=1
The equation of curve is 2x2+3y2=5
The graph for above equation is, the equation of curve and line are homogeneous so,
2x2+3y2=5(1)2
2x2+3y2=5(

3x+4y

5

)2
2x2+3y2=5(

9x2+16y2+24xy

25

)
x2−y2+24xy=0
Now,
( coefficient of x2)+ ( coefficient of y2)=1+(−1)
= 0
This implies,
angle‌AOB=90° or

π

2