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  • Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x,y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x,y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

Answers (1)

Points on the y axis and x axis are namely A(0,a), B(b,0). The midpoint of AB is P(x,y).

The x coordinate of the points is given by the addition of the x coordinates of A and B divided by 2.

\begin{aligned} &\Rightarrow x=\frac{b+0}{2}\\ &b=2 x\\ &\text { Similarly, for y coordinate }\\ &\Rightarrow y=\frac{0+a}{2}\\ &a=2 y \end{aligned}

Therefore, the coordinates of A and B are (0,2y) and (2x,0) respectively.

AB is the tangent to curve where P is the point of contact.
Slope of the line given with two points \left(x_{1,} y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ on it is $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Here \left(x_{1}, y_{1}\right)\left(x_{2}, y_{2}\right)$ are $(0,2 y)(2 x, 0)$ respectively.
Slope of the tangent AB is
\frac{0-2 y}{2 x-0}$
Hence the slope of the tangent is -y/x
Slope of the tangent curve is given by,
\\\frac{\mathrm{dy}}{\mathrm{dx}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=-\frac{\mathrm{dx}}{\mathrm{x}}$

Integrate

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

\\logy =-\log x+c$ \\logy+ $\log x=c$
using logat logb=logab.
\log x y=c$
as given curve is passing through(1,a)
Hence (1,1) will satisfy the equation of the curve(a)
Putting x=1, y=2$ in $(a)$
\\\log 1=c$ \\$c=0$
put c back in (a)
\\\log x y=0$ \\$x y=e^{0}$
$$ x y=1 $$
Hence the equation of the curve is x y=1$

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