Question

If the straight line \[ x\cos\alpha+y\sin\alpha=p \] touches the curve \[ \left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2 \] at the point \((a,b)\) on it and \[ \frac{1}{a^2}+\frac{1}{b^2}=\frac{k}{p^2}, \] then \(k=\)

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(7\)

Hint:

For tangent problems, first find the tangent equation using differentiation, then compare it with the given normal form of a straight line.

Answer 1

The Correct Option is A

Step 1: Differentiate the curve.
Given curve is \[ \left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2 \] Differentiate with respect to \(x\): \[ n\left(\frac{x}{a}\right)^{n-1}\frac{1}{a} + n\left(\frac{y}{b}\right)^{n-1}\frac{1}{b}\frac{dy}{dx}=0 \] At the point \((a,b)\), \[ \frac{x}{a}=1,\qquad \frac{y}{b}=1 \] So, \[ \frac{n}{a}+\frac{n}{b}\frac{dy}{dx}=0 \] \[ \frac{dy}{dx}=-\frac{b}{a} \]

Step 2: Write the tangent at \((a,b)\).
The tangent has slope \[ -\frac{b}{a} \] Equation of tangent at \((a,b)\) is \[ y-b=-\frac{b}{a}(x-a) \] Simplifying, \[ \frac{x}{a}+\frac{y}{b}=2 \]

Step 3: Compare with the given line.
The given line is \[ x\cos\alpha+y\sin\alpha=p \] Dividing by \(p\), \[ \frac{x\cos\alpha}{p}+\frac{y\sin\alpha}{p}=1 \] The tangent equation \[ \frac{x}{a}+\frac{y}{b}=2 \] can be written as \[ \frac{x}{2a}+\frac{y}{2b}=1 \] Comparing coefficients, \[ \frac{\cos\alpha}{p}=\frac{1}{2a} \] and \[ \frac{\sin\alpha}{p}=\frac{1}{2b} \] Thus, \[ \cos\alpha=\frac{p}{2a} \] and \[ \sin\alpha=\frac{p}{2b} \]

Step 4: Use \(\sin^2\alpha+\cos^2\alpha=1\).
\[ \left(\frac{p}{2a}\right)^2+\left(\frac{p}{2b}\right)^2=1 \] \[ \frac{p^2}{4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=1 \] \[ \frac{1}{a^2}+\frac{1}{b^2}=\frac{4}{p^2} \] Comparing with \[ \frac{1}{a^2}+\frac{1}{b^2}=\frac{k}{p^2}, \] we get \[ k=4 \]

Step 5: Final conclusion.
Hence, \[ \boxed{4} \] which corresponds to option (1).