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The slope of the tangent of the curve
\[
\left(\frac{x}{31}\right)^n+\left(\frac{y}{1209}\right)^n=2 \text { at }(31,1209) \ldots \ldots
\]
Options:
\[
\left(\frac{x}{31}\right)^n+\left(\frac{y}{1209}\right)^n=2 \text { at }(31,1209) \ldots \ldots
\]
Solution:
2321 Upvotes
Verified Answer
The correct answer is:
-39
\(\left(\frac{x}{31}\right)^n+\left(\frac{y}{1209}\right)^n=2\)
On differentiating
\(\begin{aligned}
& \Rightarrow \quad \frac{n x^{n-1}}{31^n}+\frac{n y^{n-1}}{1209^n} \frac{d y}{d x}=0 \\
& \Rightarrow \quad \frac{d y}{d x}=-\left(\frac{n x^{n-1} \times 1209^n}{n y^{n-1} \cdot 31^n}\right) \\
& \Rightarrow \quad \frac{d y}{d x}=-\left(\frac{x}{y}\right)^{n-1} \times\left(39^n\right)
\end{aligned}\)
Slope of tangent at \((31,1209)\)
\(=-\left(\frac{1}{39}\right)^{n-1} \times(39)^n=-39\)
On differentiating
\(\begin{aligned}
& \Rightarrow \quad \frac{n x^{n-1}}{31^n}+\frac{n y^{n-1}}{1209^n} \frac{d y}{d x}=0 \\
& \Rightarrow \quad \frac{d y}{d x}=-\left(\frac{n x^{n-1} \times 1209^n}{n y^{n-1} \cdot 31^n}\right) \\
& \Rightarrow \quad \frac{d y}{d x}=-\left(\frac{x}{y}\right)^{n-1} \times\left(39^n\right)
\end{aligned}\)
Slope of tangent at \((31,1209)\)
\(=-\left(\frac{1}{39}\right)^{n-1} \times(39)^n=-39\)
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