TS EAMCET 10 May 2024 Shift 1 Solved Paper

Let the variable point be P(h,k).
The distance from P to the point (4,3) can be expressed as:
√(h−4)2+(k−3)2
The perpendicular distance from P to the line x+2y−1=0 is given by:
|‌

h+2k−1

√12+22

|=|‌

h+2k−1

√5

|
According to the problem, these two distances are equal:
√(h−4)2+(k−3)2=|‌

h+2k−1

√5

|
Squaring both sides, we get:
(h−4)2+(k−3)2=‌

(h+2k−1)2

5

Expanding both sides:
h2−8h+16+k2−6k+9=‌

h2+4hk+4k2−2h−4k+1

5

Multiply through by 5 to eliminate the fraction:
5(h2−8h+16+k2−6k+9)=h2+4hk+4k2−2h−4k+1
Simplify and collect like terms:
5h2+5k2−40h−30k+125=h2+4hk+4k2−2h−4k+1

Rearranging gives:
4h2+k2−4hk−38h−26k+124=0
Finally, replace (h,k) with (x,y) :
4x2+y2−4xy−38x−26y+124=0