Let the tangent at any point P on a curve passing through the points (1, 1) and [((1)/(10)…

Question

Let the tangent at any point P on a curve passing through the points (1, 1) and $$\left(\frac{1}{10}, 100\right)$$, intersect positive $x$-axis and $y$-axis at the points A and B respectively. If $$\mathrm{PA}: \mathrm{PB}=1: k$$

Accepted Answer

Let the equation of tangent to the curve at $(x,y)$.

Whose slope is $\frac{dy}{dx}$ is

$\begin{gathered} Y-y=\frac{dy}{dx}(X-x) \ \ Y=0\Rightarrow X=x-\frac{y}{(\frac{dy}{dx})}t \ \ \Rightarrow \alpha =x-\frac{y}{(\frac{dy}{dx})} \end{gathered}$

and putting $X=0\Rightarrow Y=y-x\frac{dy}{dx}$

$\Rightarrow \beta =y-x\frac{dy}{dx}$

$\because P$ divides $AB$ in $1:k$

$x=\frac{k\alpha +0}{k+1}y=\frac{k	imes 0+\beta }{k+1}$

$\begin{gathered} \Rightarrow x(k+1)=k(x-\frac{y}{\frac{dy}{dx}}) \ \Rightarrow xk+x=xk-\frac{yk}{\frac{dy}{dx}} \ \Rightarrow x\frac{dy}{dx}=-yk \end{gathered}$

$\begin{gathered} \int \frac{dy}{y}=-k	imes \int \frac{1}{x}dx \ \Rightarrow \log y=-k\log x+\log C \ \log y	imes {x}^{k}=\log C \ \Rightarrow y{x}^{k}=C \ x=1,y=1\Rightarrow c=1 \ y{x}^{k}=1 \end{gathered}$

Putting $x=\frac{1}{10},y=100\Rightarrow 100	imes {(\frac{1}{100})}^{k}=1\Rightarrow k=2$

so $y{x}^{2}=1$ or $y=\frac{1}{{x}^{2}}$

Now ${e}^{\frac{dy}{dx}}=kx+\frac{k}{2}$

$\Rightarrow \frac{dy}{dx}={\log }_{e}(kx+\frac{k}{2})={\log }_{e}(2x+1)$

On integrating

$\begin{gathered} y=\int 1\cdot {\log }_{e}(2x+1)dx \ =x{\log }_{e}(2x+1)-\int \frac{1	imes 2}{2x+1}	imes xdx \ =x{\log }_{e}(2x+1)-\int 1-\frac{1}{2x+1}dx \ y=x{\log }_{e}(2x+1)-x+\frac{1}{2}{\log }_{e}(2x+1)+c \end{gathered}$

Put $x=0,y=k=2\Rightarrow c=2$ putting $x=1$

$\begin{gathered} y(1)={\log }_{e}3-1+\frac{1}{2}{\log }_{e}3+2=\frac{3}{2}{\log }_{e}3-1+2 \ \Rightarrow 4y(1)=6{\log }_{e}3+4 \ \Rightarrow 4y(1)-6{\log }_{e}3=4 \end{gathered}$

Solutions · 1