Question

The sum of the intercepts made by a tangent drawn to the curve \((x/a)^n + (y/b)^n = 2\) at \((a,b)\) on the coordinate axes is

• A. \(a+b\)
• B. \(a^2 + b^2\)
• C. \(2(a-b)\)
• D. \(2(a+b)\)

Hint:

For a curve \( (x/a)^n + (y/b)^n = \text{const} \), tangent at point \((a,b)\) in linear form gives intercepts; sum = sum of intercepts on axes.

Answer 1

The Correct Option is D

Step 1: Differentiate the curve implicitly.
For \((x/a)^n+(y/b)^n=2\), differentiating both sides with respect to \(x\) gives
\[ \frac{n}{a}\left(\frac{x}{a}\right)^{n-1} + \frac{n}{b}\left(\frac{y}{b}\right)^{n-1} y' = 0 \]

Step 2: Evaluate the slope at (a, b).
At \((a,b)\), both bracketed terms equal 1, so \(\frac{n}{a}+\frac{n}{b}y'=0\), giving \(y'=-\frac{b}{a}\).

Step 3: Write the tangent line and find intercepts.
Tangent: \(y-b=-\frac{b}{a}(x-a)\), which simplifies to \(\frac{x}{a}+\frac{y}{b}=2\). Setting \(y=0\) gives \(x=2a\); setting \(x=0\) gives \(y=2b\).

Step 4: Final conclusion.
Sum of intercepts \(=2a+2b=\) \[ \boxed{2(a+b)} \]