Application of Derivatives
Tangents and Normals
IIT-JEE 2001
Grade 12
Question:
For the curve $x = t^2 + 3t - 8$, $y = 2t^2 - 2t - 5$, the slope of the tangent at the point $(2, -1)$ is:
(1) $\frac{6}{7}$
(2) $\frac{7}{6}$
(3) $\frac{22}{7}$
(4) $\frac{6}{11}$
Step-by-Step Solution
Key Concept: First find the value of the parameter $t$ that gives the point $(2, -1)$ by solving the two coordinate equations together, then form $\frac{dy/dt}{dx/dt}$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)