MHT CET · Maths · Application of Derivatives
The curve \(\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2, n \in N\) touches the line at the point \((a, b)\), then the equation of the line is
- A \(\frac{x}{a}-\frac{y}{b}=2\)
- B \(\frac{x}{a}+\frac{y}{2 b}=1\)
- C \(\frac{x}{a}+\frac{y}{b}=1\)
- D \(\frac{x}{a}+\frac{y}{b}=2\)
Answer & Solution
Correct Answer
(D) \(\frac{x}{a}+\frac{y}{b}=2\)
Step-by-step Solution
Detailed explanation
Slope of the line \(=\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \((a, b)\)
\(\begin{aligned} & =\frac{-n\left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a}}{n \cdot\left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b}} \text { at }(a, b) \\ & =\frac{-b}{a}\end{aligned}\)
now equation of the straight line
\(\begin{aligned} & y-b=-\frac{b}{a}(x-a) \\ & \Rightarrow a y-a b=-b x+a b \\ & \Rightarrow b x+a y=2 a b \\ & \Rightarrow \frac{x}{a}+\frac{y}{b}=2\end{aligned}\)
\(\begin{aligned} & =\frac{-n\left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a}}{n \cdot\left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b}} \text { at }(a, b) \\ & =\frac{-b}{a}\end{aligned}\)
now equation of the straight line
\(\begin{aligned} & y-b=-\frac{b}{a}(x-a) \\ & \Rightarrow a y-a b=-b x+a b \\ & \Rightarrow b x+a y=2 a b \\ & \Rightarrow \frac{x}{a}+\frac{y}{b}=2\end{aligned}\)
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