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# How to solve this problem- Tangent to any point P of curve meets x-axi in T, Such that OP = PT (O is origin), then curve is

Tangent to any point P of curve meets x-axi in T, Such that OP = PT (O is origin),  then curve is

• Option 1)

$xy =c$

• Option 2)

$xy ^2=c$

• Option 3)

$x^2y =c$

• Option 4)

$x +y =c$

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As we have learnt,

Equation of tangent at a point (x y) -

$Y-y= \frac{dy}{dx}\left ( X-x \right )$

-

Equation of tangent will be

$Y-y= \frac{dy}{dx}\left ( X-x \right )$ at P (x, y)

For T, Y = 0

$\\*\Rightarrow -y = \frac{dy}{dx}(X- x) \Rightarrow X = x -y\frac{\mathrm{d} x}{\mathrm{d} y} \\*\therefore T(x - y\frac{\mathrm{d} x}{\mathrm{d} y}, 0)$

$\because OP = PT \Rightarrow OP^2 = PT^2 \Rightarrow x^2 + y^2 = y^2\left(\frac{dx}{dy}\right)^2 + y^2 \\*\Rightarrow \frac{dx}{dy} = \frac{x}{y}\;or\;\frac{dx}{dy} = -\frac{x}{y}$

On Integrating, we get

either $y=cx\;or\;xy = c$

Option 1)

$xy =c$

Option 2)

$xy ^2=c$

Option 3)

$x^2y =c$

Option 4)

$x +y =c$

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