AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Consider the figure shown below.Let S=3x2+2y2−5
The equation of pair of tangent drawn to the ellipse from the point (1,2) is,
SS1=T2
Therefore,
(3x2+2y2−5)(3(1)2+2(2)2−5) =(3x(1)+2y(2)−5)2
(3x2+2y2−5)6 =(3x+4y−5)2
9x2−4y2−24xy+30x +40y−55=0
Hence,
a=9
h=−12
b=−4
The angle between the line is calculated as,
tan‌θ=

2√h2−ab

a+b

=

2√144+36

5

=

12√5

5

θ=tan−1(

12√5

5

)