NIMCET 2025 — Mathematics PYQ

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Question

NIMCET 2025 — Mathematics PYQ

NIMCET | Mathematics | 2025

The slope of the normal line to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is:

Choose the correct answer:

Explanation

Solution

Step 1: Solve for $t$

At $(2, -1)$ :

x = t^2 + 3t - 8 = 2 \implies t^2 + 3t - 10 = 0 \implies (t+5)(t-2) = 0

y = 2t^2 - 2t - 5 = -1 \implies 2t^2 - 2t - 4 = 0 \implies t^2 - t - 2 = 0 \implies (t-2)(t+1) = 0

Common value is $t = 2$ .

Step 2: Find Derivatives

\frac{dx}{dt} = 2t + 3

\frac{dy}{dt} = 4t - 2

Step 3: Slope of Tangent ( $m_t$ )

At $t = 2$ :

\frac{dx}{dt} = 2(2) + 3 = 7

\frac{dy}{dt} = 4(2) - 2 = 6

m_t = \frac{dy/dt}{dx/dt} = \frac{6}{7}

Step 4: Slope of Normal ( $m_n$ )

m_n = -\frac{1}{m_t} = -\frac{7}{6}

Final Answer:

The slope of the normal is $-\frac{7}{6}$ . (Option D)