If the straight line x cos alpha + y sin alph | vedclass

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(A) The given curve is $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$. At the point $(a, b)$,the curve satisfies the equation $(\frac{a}{a})^n + (\frac{b}{b}

Vedclass · Accepted Answer

$4$

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(A) The given curve is $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$. At the point $(a, b)$,the curve satisfies the equation $(\frac{a}{a})^n + (\frac{b}{b})^n = 1^n + 1^n = 2$,which is true for any $n$. Differentiating the curve equation with respect to $x$: $\frac{n}{a^n} x^{n-1} + \frac{n}{b^n} y^{n-1} \frac{dy}{dx} = 0$. At the point $(a, b)$,the slope of the tangent is: $(\frac{dy}{dx})_{(a, b)} = -\frac{b^{n-1}}{a^{n-1}} \cdot \frac{a^n}{b^n} = -\frac{b}{a}$. The equation of the tangent at $(a, b)$ is $y - b = -\frac{b}{a}(x - a)$,which simplifies to $bx + ay = 2ab$. Comparing this with $x \cos \alpha + y \sin \alpha = p$,we identify $\cos \alpha = \frac{b}{p}$ and $\sin \alpha = \frac{a}{p}$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have $\frac{b^2}{p^2} + \frac{a^2}{p^2} = 1$,so $a^2 + b^2 = p^2$. Given $\frac{1}{a^2} + \frac{1}{b^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{p^2}{a^2 b^2} = \frac{k}{p^2}$. From $p = 2ab$,we have $p^2 = 4a^2 b^2$,so $a^2 b^2 = \frac{p^2}{4}$. Substituting this into the expression: $\frac{p^2}{p^2/4} = 4 = \frac{k}{p^2} \cdot p^2 = k$. Thus,$k = 4$.

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

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