If β is an angle between the normals drawn to the curve x2+3y2=9 at the points (3‌cos‌θ,√3sin‌θ) and (−3sin‌θ,√3‌cos‌θ),θ∈(0,‌
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=‌