The slope of the normal line to the curve x=t²+3t-8, \;\; y=2t²-2t-5 at the point (2,-1) is:

Correct Answer is : −76-7/6−67

Correct Answer is : −76-7/6−67

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NIMCET 2025 Paper

Section: Mathematics

Question : 37 of 120

Marks: +1, -0

x=t^2+3t-8, \;\; y=2t^2-2t-5

(2,-1)

Solution:

To find the normal slope at

(2,-1)

, first find the value of

t

that gives this point.

Solve:

t^2+3t-8=2 \Rightarrow t^2+3t-10=0

This factors:

(t+5)(t-2)=0 \Rightarrow t=2 \; \text{(since it must give } \; y=-1)

Now compute derivatives:

\frac{dx}{dt}=2t+3, \;\; \frac{dy}{dt}=4t-2 .

t=2

\frac{dx}{dt}=7, \;\; \frac{dy}{dt}=6

So the slope of the tangent is:

\frac{dy}{dx}= \frac{6}{7}

Thus, the slope of the normal is the negative reciprocal:

m_{\text{normal}} = -\frac{7}{6}

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