Question

Prove that the condition $x\cos \alpha +y\sin \alpha =p$ touches the curve $x^m{y}^n=a^{m+n}$ is $p^A{m}^n{n}^m=A^A{a}^A.{\cos }^n\alpha {\sin }^m\alpha$ is then $A=$

• A. $m+n$
• B. $m-n$
• C. $n-m$
• D. $m^2-n^2$

Answer 1

The correct option is A

$m+n$

Explanation for the correct answer:

Finding the value of $A$:

Given,

$x\cos \alpha +y\sin \alpha =p\to \left(i\right)$

$x^m{y}^n=a^{m+n}\to \left(ii\right)$

Differentiating equation $\left(i\right)$ with respect to '$x$'

$$\begin{gathered} \Rightarrow \cos \alpha +\frac{dy}{dx}\sin \alpha =0 \\ \Rightarrow \frac{dy}{dx}=-\frac{\cos \alpha }{\sin \alpha } \\ \Rightarrow \frac{dy}{dx}=-\cot \left(\alpha \right)\to \left(iii\right) \end{gathered}$$

Differentiating equation $\left(ii\right)$ with respect to '$x$'

$\Rightarrow m{x}^{m-1}{y}^n+x^{m}n.y^{n-1}\frac{dy}{dx}=0$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=-\frac{m{x}^{m-1}{y}^n}{x^{m}n.y^{n-1}} \\ \Rightarrow \frac{dy}{dx}=-\frac{my}{nx}\to \left(iv\right) \end{gathered}$$

Equating equation $\left(iii\right)\text{and}\left(iv\right)$ we get,

$\Rightarrow \cot \left(\alpha \right)=\frac{my}{nx}$

Divide equation $\left(i\right)$ by $\sin \left(\alpha \right)$

$$\begin{gathered} \Rightarrow x\frac{\cos \alpha }{\sin \alpha }+y=\frac{p}{\sin \alpha } \\ \Rightarrow x\cot \left(\alpha \right)+y=\frac{p}{\sin \alpha } \\ \Rightarrow x\frac{my}{nx}+y=\frac{p}{\sin \alpha } \\ \Rightarrow y\left(\frac{m}{n}+1\right)=\frac{p}{\sin \alpha } \\ \Rightarrow y\left(\frac{m+n}{n}\right)=\frac{p}{\sin \alpha } \\ \Rightarrow y\left(m+n\right)\sin \alpha =p.n \end{gathered}$$

Raising the equation to the power $m$

$$\begin{gathered} \Rightarrow {\left(y\left(m+n\right)\sin \alpha \right)}^m={\left(p.n\right)}^m \\ \Rightarrow y^m{\left(m+n\right)}^m{\sin }^m\alpha =p^m.n^m\to \left(1\right) \end{gathered}$$

Divide equation $\left(i\right)$ by $\cos \left(\alpha \right)$

$$\begin{gathered} \Rightarrow x+y\frac{\sin \alpha }{\cos \alpha }=\frac{p}{\cos \left(\alpha \right)} \\ \Rightarrow x+y\tan \left(\alpha \right)=\frac{p}{\cos \left(\alpha \right)} \\ \Rightarrow x+y\frac{nx}{my}=\frac{p}{\cos \left(\alpha \right)}\left[\because \cot \left(\alpha \right)=\frac{my}{nx}\Rightarrow \tan \left(\alpha \right)=\frac{1}{\cot \left(\alpha \right)}=\frac{nx}{my}\right] \\ \Rightarrow x\left(1+\frac{n}{m}\right)=\frac{p}{\cos \left(\alpha \right)} \\ \Rightarrow x\left(\frac{m+n}{m}\right)=\frac{p}{\cos \left(\alpha \right)} \\ \Rightarrow x\left(m+n\right)\cos \left(\alpha \right)=p.m \end{gathered}$$

Raising the equation to the power $n$

$$\begin{gathered} \Rightarrow {\left(x\left(m+n\right)\cos \alpha \right)}^n={\left(p.m\right)}^n \\ \Rightarrow y^n{\left(m+n\right)}^n{\cos }^n\alpha =p^n.m^n\to \left(2\right) \end{gathered}$$

Multiply equation $1\text{and}2$, we get,

$$\begin{gathered} \Rightarrow \left(y^m{\left(m+n\right)}^m{\sin }^m\alpha \right)\times \left(y^n{\left(m+n\right)}^n{\cos }^n\alpha \right)=\left(p^n.m^n\right)\left(p^m.n^m\right) \\ \Rightarrow y^{m+n}{\left(m+n\right)}^{m+n}{\sin }^m\alpha {\cos }^n\alpha =p^{m+n}.m^n.n^m \end{gathered}$$

Equating this with the given equation $p^A{m}^n{n}^m=A^A{a}^A.{\cos }^n\alpha {\sin }^m\alpha$, we get: $A=m+n$

Hence, the correct answer is option (A).