Question

For the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$, ($n \in N$ & $n>1$) the line $\frac{x}{a}+\frac{y}{b}=2$ is

• A. a normal for all values of n
• B. a normal for only values of n more than Max\{a,b\}
• C. a tangent for all values of n
• D. a tangent for only values of n more than Min\{a,b\}

Hint:

To prove that a line is tangent to a curve at a specific point, you must show two things: (1) the point lies on both the line and the curve, and (2) the slope of the line is equal to the derivative of the curve's function evaluated at that point.

Answer 1

The Correct Option is C

Step 1: Check for a point of intersection.
Let's see if there is an obvious point that lies on both the curve and the line. Consider the point $P(a,b)$. For the line: $\frac{a}{a} + \frac{b}{b} = 1+1=2$. So, $P(a,b)$ is on the line. For the curve: $\left(\frac{a}{a}\right)^n + \left(\frac{b}{b}\right)^n = 1^n + 1^n = 1+1=2$. So, $P(a,b)$ is on the curve for all $n$. Thus, the line and the curve intersect at the point $(a,b)$.

Step 2: Find the slope of the tangent to the curve at $(a,b)$.
We use implicit differentiation on the curve's equation: $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$. \[ \frac{d}{dx}\left((\frac{x}{a})^n + (\frac{y}{b})^n\right) = \frac{d}{dx}(2). \] \[ n\left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a} + n\left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0. \] Now, solve for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = - \frac{n(x/a)^{n-1} \cdot (1/a)}{n(y/b)^{n-1} \cdot (1/b)} = -\frac{b}{a} \left(\frac{x/a}{y/b}\right)^{n-1}. \] Evaluate this slope at the point $(a,b)$: \[ m_{\text{tangent}} = -\frac{b}{a} \left(\frac{a/a}{b/b}\right)^{n-1} = -\frac{b}{a} (1)^{n-1} = -\frac{b}{a}. \]

Step 3: Find the slope of the given line.
The equation of the line is $\frac{x}{a}+\frac{y}{b}=2$. Rearranging into slope-intercept form ($y=mx+c$): \[ \frac{y}{b} = 2 - \frac{x}{a} \implies y = 2b - \frac{b}{a}x. \] The slope of the line is $m_{\text{line}} = -\frac{b}{a}$.

Step 4: Conclude based on the results.
We found that the line passes through the point $(a,b)$ on the curve. We also found that the slope of the line is equal to the slope of the tangent to the curve at that same point. These two conditions together mean that the line is tangent to the curve at the point $(a,b)$. Since this holds true for any integer $n>1$, the line is a tangent for all values of $n$.