AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

The equation of the line that passes through P(a,2) and a≠0 making an angle 45∘ with positive X-axis is,
‌

x−a

1 √2

=‌

y−2

1 √2

=r
x=a+‌

r

√2

and y=2+‌

r

√2

So
r=−a√2
For point C
PB‌‌=√4+4
‌‌=2√2
And,
PC‌‌=√a2+a2
 ‌‌=a√2
From the points A and D
‌

(a+‌ r √2 )2

9

+‌

(2+‌ r √2 )2

4

=1
4(a+‌

r

√2

)2+9(2+‌

r

√2

)2=36
The roots are given by,
r1r2=‌

4a2

13 2

 =‌

8a2

13

 =(PA)(PD)
This implies,
PA,PB,PC,PD are in G.P so
(PA)(PD)‌‌=(PB)(PC)
  ‌

8a2

13

‌‌=4a
   2a‌‌=13