AP EAPCET 22 May 2025 Shift 2 Paper

Step 1: Find the Derivative and Slope of Normal
The curve is x2+3y2=9.
Differentiate both sides with respect to x :
2x+3⋅2y⋅y′=0
Solve for y′ :
2x+6yy′=0⟹6yy′=−2x⟹y′=−‌

2x

6y

=−‌

x

3y

The slope of the tangent is y′=−‌

x

3y

, so the slope of the normal is −‌

1

y′

=‌

3y

x

.
Step 2: Find the Slope of Normals at Given Points
First point: (3‌cos‌θ,√3sin‌θ)
Slope at this point:
m1=‌

3⋅√3sin‌θ

3‌cos‌θ

=√3‌tan‌θ
Second point: (−3sin‌θ,√3‌cos‌θ)
Slope at this point:
m2=‌

3⋅√3‌cos‌θ

−3sin‌θ

=−√3‌cot‌θ
Step 3: Find the Angle Between Normals ( β )
Use the formula for the tangent of the angle between two lines:
tan‌β=|‌

m1−m2

1+m1m2

|
‌tan‌β=|‌

√3‌tan‌θ−(−√3‌cot‌θ)

1+(√3‌tan‌θ)(−√3‌cot‌θ)

|
‌tan‌β=|‌

√3(tan‌θ+cot‌θ)

1−3

|
‌tan‌β=|‌

√3( sin‌θ cos‌θ + cos‌θ sin‌θ )

−2

|
‌tan‌β=‌

√3

2

⋅‌

1

sin‌θ‌cos‌θ

Recall 2sin‌θ‌cos‌θ=sin‌2θ, so:
tan‌β=‌

√3

sin‌2θ

This means ‌

1

cot‌β

=‌

√3

sin‌2θ

or √3‌cot‌β=sin‌2θ.