Question

Find the condition that the line $Ax+By=1$ may be a normal to the curve $a^{n-1}y=x^n.$

• A. $a^{n}B{(B^2+n{A}^2)}^n=A^n{n}^n.$
• B. $a^{n-1}B{(B^2+n{A}^2)}^{n-1}=A^n{n}^n.$
• C. $a^{n}B{(B^2+n{A}^2)}^{n-1}=A^n{n}^n.$
• D. $a^{n-1}B{(B^2-n{A}^2)}^{n-1}=A^n{n}^n.$

Answer 1

Answer: (B)

$a^{n-1}B{(B^2+n{A}^2)}^{n-1}=A^n{n}^n.$

Given, $a^{n-1}y=x^n$

$\therefore \frac{dy}{dx}=n\frac{x^{n-1}}{a^{n-1}}=n\frac{x^{n-1}}{a^{n-1}}\cdot \frac{1}{x}=n\frac{y}{x}.$

$\therefore$ Normal is: $Y-y=-\frac{1}{dy/dx}(X-x)=-\frac{x}{ny}(X-x)$

$\therefore Xx+Yny=n{y}^2+x^2.$

Compare with $AX+BY=1$

$\therefore \frac{X}{A}=\frac{ny}{B}=\frac{n{y}^2+x^2}{1}=k,$ say.

$\therefore x=Ak,y=\left(Bk/n\right)$ and $n{y}^2+x^2=k$ or $k^2\left[(B^2/n+A^2)\right]=k$

$\therefore k=\frac{n}{B^2+n{A}^2}$ ..(1)

Now $a^{n-1}y=x^n.$

Put for $x$ and $y$.

$a^{n-1}y\cdot \frac{Bk}{n}=A^n{k}^n$ $\Rightarrow a^{n-1}B=n{A}^n{k}^{n-1}$

$\Rightarrow a^{n-1}B=n{A}^n\cdot {\left(\frac{n}{B^2+n{A}^2}\right)}^{n-1},$ by (1)

$\Rightarrow a^{n-1}B{(B^2+n{A}^2)}^{n-1}=A^n{n}^n.$

Above is the required condition.