AP EAPCET 24 May 2025 Shift 1 Paper

Given equation is
2x2+2hxy+2y2−x+y=1
The angle θ between the lines is tan−1(‌

3

4

)=0⇒tan‌θ=‌

3

4

. The angle θ between the pair of lines ax2+2hxy+by2+2gx+2fy+c=0 is
tan‌θ=|‌

2√h2−ab

a+b

|
Here a=2,b=2
tan‌θ‌=|‌

2√h2−(2)(2)

2+2

|=|‌

2√h2−4

4

|
‌=|‌

√h2−4

2

|
‌⇒‌

3

4

=‌

√h2−4

2

⇒‌

h2−4

4

=‌

9

16

‌⇒h2−4=‌

9

16

×4=‌

9

4

‌⇒h2=‌

9

4

+4=‌

25

4

‌⇒h=±√‌

25

4

=±‌

5

2

Since, h is a positive rational number,
h=‌

5

2

Thus, the given equation is
‌2x2+2hxy+2y2−x+y−1=0
‌‌‌2x2+2⋅‌

5

2

xy+2y2−x+y−1=0
‌‌‌2x2+5xy+2y2−x+y+1=0
‌‌‌=F(x,y)
Now, ‌

∂F

∂x

=4x+5y−1=0,
‌

∂F

∂y

=4y+5x+1=0
Solving these two equations, we get x=−1,y=1
So, the point of intersection is (−1,1).