If length of tangent at any point on the curve y = f(x) intercepted be

1.	If length of tangent at any point on the curve y = f(x) intercepted between the point and the X - axis is of length 1. Find the equation of the curve.
Answer» $\large\underline{\sf{Solution-}}$ Given that, Length of tangent at any point on the curve y = f(x), intercepted between the point and the X - axis is of length 1 unit. We KNOW, $\red{\rm :\longmapsto\:Length \: of \: tangent = \bigg \|y \sqrt{1 + {\bigg(\dfrac{dx}{dy} \bigg) }^{2} } \bigg\|}$ So, given that Length of tangent = 1 $\rm :\longmapsto\:\bigg \|y \sqrt{1 + {\bigg(\dfrac{dx}{dy} \bigg) }^{2} } \bigg\| = 1$ On squaring both sides, we get $\rm :\longmapsto\: {y}^{2}\bigg(1 + {\bigg(\dfrac{dx}{dy} \bigg) }^{2} \bigg) = 1$ $\rm :\longmapsto\:1 + {\bigg(\dfrac{dx}{dy}\bigg) }^{2} = \dfrac{1}{ {y}^{2} }$ $\rm :\longmapsto\: {\bigg(\dfrac{dx}{dy}\bigg) }^{2} = \dfrac{1}{ {y}^{2} } - 1$ $\rm :\longmapsto\: {\bigg(\dfrac{dx}{dy}\bigg) }^{2} = \dfrac{1 - {y}^{2} }{ {y}^{2} }$ $\rm :\longmapsto\: {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = \dfrac{{y}^{2} }{1 - {y}^{2} }$ $\bf\implies \:\dfrac{dy}{dx} \: = \: \pm \: \dfrac{y}{ \sqrt{1 - {y}^{2} } }$ $\rm :\longmapsto\: \dfrac{ \sqrt{1 - {y}^{2} } }{y}dy \: = \: \pm \: \: dx$ On integrating both sides, we get $\rm :\longmapsto\:\displaystyle\int \dfrac{ \sqrt{1 - {y}^{2} } }{y}dy \: = \: \pm \: \: \displaystyle\int dx$ $\rm :\longmapsto\:\displaystyle\int \dfrac{y \sqrt{1 - {y}^{2} } }{ {y}^{2} }dy \: = \: \pm \: x + c$ $\rm :\longmapsto\:\displaystyle\int \dfrac{\sqrt{1 - {y}^{2} } }{ {y}^{2} }ydy \: = \: \pm \: x + c$ To integrate LHS, we use METHOD of SUBSTITUTION Put $\red{\rm :\longmapsto\: \sqrt{1 - {y}^{2} } = t}$ $\red{\rm :\longmapsto\:1 - {y}^{2} = {t}^{2}}$ $\red{\rm :\longmapsto\: - 2ydy = 2tdt}$ $\red{\rm :\longmapsto\: ydy \: = \: - \: tdt}$ So, above integral can be rewritten as, $\rm :\longmapsto\:\displaystyle\int \frac{t}{1 - {t}^{2} } ( - t \: dt) = \: \pm \: x \: + \: c$ $\rm :\longmapsto\:\displaystyle\int \frac{ - {t}^{2} }{1 - {t}^{2} } \: dt = \: \pm \: x \: + \: c$ $\rm :\longmapsto\:\displaystyle\int \frac{1 - {t}^{2} - 1}{1 - {t}^{2} } \: dt = \: \pm \: x \: + \: c$ $\rm :\longmapsto\:\displaystyle\int \frac{1 - {t}^{2}}{1 - {t}^{2} } \: dt - \displaystyle\int \frac{1}{1 - {t}^{2} }dt = \: \pm \: x \: + \: c$ $\rm :\longmapsto\:\displaystyle\int 1 \: dt + \displaystyle\int \frac{1}{{t}^{2} - 1}dt = \: \pm \: x \: + \: c$ We know, $\boxed{ \rm{ \displaystyle\int \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a}log\bigg \| \frac{x - a}{x + a} \bigg\| + c}}$ $\rm :\longmapsto\:t + \dfrac{1}{2}log\bigg \| \dfrac{t - 1}{t + 1} \bigg\| = \: \pm \: x + c$ $\rm :\longmapsto\: \sqrt{1 - {y}^{2} } + \dfrac{1}{2}log\bigg \| \dfrac{ \sqrt{1 - {y}^{2} } - 1}{ \sqrt{1 - {y}^{2} } + 1} \bigg\| = \: \pm \: x + c$ is the required curve.

If length of tangent at any point on the curve y = f(x) intercepted between the point and the X - axis is of length 1. Find the equation of the curve.

Answer»

$\large\underline{\sf{Solution-}}$

Given that,

Length of tangent at any point on the curve y = f(x), intercepted between the point and the X - axis is of length 1 unit.

We KNOW,

$\red{\rm :\longmapsto\:Length \: of \: tangent = \bigg |y \sqrt{1 + {\bigg(\dfrac{dx}{dy} \bigg) }^{2} } \bigg|}$

So, given that Length of tangent = 1

$\rm :\longmapsto\:\bigg |y \sqrt{1 + {\bigg(\dfrac{dx}{dy} \bigg) }^{2} } \bigg| = 1$

On squaring both sides, we get

$\rm :\longmapsto\: {y}^{2}\bigg(1 + {\bigg(\dfrac{dx}{dy} \bigg) }^{2} \bigg) = 1$

$\rm :\longmapsto\:1 + {\bigg(\dfrac{dx}{dy}\bigg) }^{2} = \dfrac{1}{ {y}^{2} }$

$\rm :\longmapsto\: {\bigg(\dfrac{dx}{dy}\bigg) }^{2} = \dfrac{1}{ {y}^{2} } - 1$

$\rm :\longmapsto\: {\bigg(\dfrac{dx}{dy}\bigg) }^{2} = \dfrac{1 - {y}^{2} }{ {y}^{2} }$

$\rm :\longmapsto\: {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = \dfrac{{y}^{2} }{1 - {y}^{2} }$

$\bf\implies \:\dfrac{dy}{dx} \: = \: \pm \: \dfrac{y}{ \sqrt{1 - {y}^{2} } }$

$\rm :\longmapsto\: \dfrac{ \sqrt{1 - {y}^{2} } }{y}dy \: = \: \pm \: \: dx$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int \dfrac{ \sqrt{1 - {y}^{2} } }{y}dy \: = \: \pm \: \: \displaystyle\int dx$

$\rm :\longmapsto\:\displaystyle\int \dfrac{y \sqrt{1 - {y}^{2} } }{ {y}^{2} }dy \: = \: \pm \: x + c$

$\rm :\longmapsto\:\displaystyle\int \dfrac{\sqrt{1 - {y}^{2} } }{ {y}^{2} }ydy \: = \: \pm \: x + c$

To integrate LHS, we use METHOD of SUBSTITUTION

Put

$\red{\rm :\longmapsto\: \sqrt{1 - {y}^{2} } = t}$

$\red{\rm :\longmapsto\:1 - {y}^{2} = {t}^{2}}$

$\red{\rm :\longmapsto\: - 2ydy = 2tdt}$

$\red{\rm :\longmapsto\: ydy \: = \: - \: tdt}$

So, above integral can be rewritten as,

$\rm :\longmapsto\:\displaystyle\int \frac{t}{1 - {t}^{2} } ( - t \: dt) = \: \pm \: x \: + \: c$

$\rm :\longmapsto\:\displaystyle\int \frac{ - {t}^{2} }{1 - {t}^{2} } \: dt = \: \pm \: x \: + \: c$

$\rm :\longmapsto\:\displaystyle\int \frac{1 - {t}^{2} - 1}{1 - {t}^{2} } \: dt = \: \pm \: x \: + \: c$

$\rm :\longmapsto\:\displaystyle\int \frac{1 - {t}^{2}}{1 - {t}^{2} } \: dt - \displaystyle\int \frac{1}{1 - {t}^{2} }dt = \: \pm \: x \: + \: c$

$\rm :\longmapsto\:\displaystyle\int 1 \: dt + \displaystyle\int \frac{1}{{t}^{2} - 1}dt = \: \pm \: x \: + \: c$

We know,

$\boxed{ \rm{ \displaystyle\int \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a}log\bigg | \frac{x - a}{x + a} \bigg| + c}}$

$\rm :\longmapsto\:t + \dfrac{1}{2}log\bigg | \dfrac{t - 1}{t + 1} \bigg| = \: \pm \: x + c$

$\rm :\longmapsto\: \sqrt{1 - {y}^{2} } + \dfrac{1}{2}log\bigg | \dfrac{ \sqrt{1 - {y}^{2} } - 1}{ \sqrt{1 - {y}^{2} } + 1} \bigg| = \: \pm \: x + c$

is the required curve.

If length of tangent at any point on the curve y = f(x) intercepted between the point and the X - axis is of length 1. Find the equation of the curve.

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