Tangents And Normals
The slope of the tangent to the curve x=t2+3t−8,y=2t2−2t−5 at the point (2,−1) is:
Select the correct option:
Solution
6/7
-
Find the parameter t for point (2,−1): x=t2+3t−8=2⟹t2+3t−10=0⟹(t+5)(t−2)=0. So t=2 or −5. y=2t2−2t−5=−1⟹2t2−2t−4=0⟹2(t2−t−2)=0⟹2(t−2)(t+1)=0. So t=2 or −1. The common value is t=2.
-
Calculate derivatives: dx/dt=2t+3. dy/dt=4t−2.
-
Slope dy/dx: dy/dx=(dy/dt)/(dx/dt)=(4t−2)/(2t+3). At t=2: (8−2)/(4+3)=6/7.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- mathematics
- Chapter
- limit, continuity and differentiability
- Topic
- tangents and normals
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
6/7
-
Find the parameter t for point (2,−1): x=t2+3t−8=2⟹t2+3t−10=0⟹(t+5)(t−2)=0. So t=2 or −5. y=2t2−2t−5=−1⟹2t2−2t−4=0⟹2(t2−t−2)=0⟹2(t−2)(t+1)=0. So t=2 or −1. The common value is t=2.
-
Calculate derivatives: dx/dt=2t+3. dy/dt=4t−2.
-
Slope dy/dx: dy/dx=(dy/dt)/(dx/dt)=(4t−2)/(2t+3). At t=2: (8−2)/(4+3)=6/7.
This easy difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of tangents and normals. It appeared in the 2025 exam.
Looking for more practice? Explore all mathematics questions or browse limit, continuity and differentiability questions on RankGuru.