# If the distance between a tangent to the parabola y^2 = 4x and a parallel normal to the same parabola is 2sqrt2, then possible values of gradient of either of them are?

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dk_ch Share
Feb 12, 2018

Parametrically let us consider that $\left({t}^{2} , 2 t\right)$ represents the coordinates of any point $P$ on the given parabola,${y}^{2} = 4 x$.

Differentiating the equation of the parabola w r to $x$ we get

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 4$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{y}$

So slope of the tangent at $P$ and also that of a normal at $Q$ parallel to this tangent will be given by

$m = {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{\text{("t^2,2t")}} = \frac{2}{2 t} = \frac{1}{t}$

Hence the tangent at the point $Q$ will have slope $m ' = - t$. So applying reverse logic we can say that the coordinates of $Q$ will be $\left(\frac{1}{t} ^ 2 , - \frac{2}{t}\right)$

Now equation of the tangent at $Q$ will be

$y - 2 t = m \left(x - {t}^{2}\right)$

$\implies y - 2 t = \frac{1}{t} \left(x - {t}^{2}\right)$

$\implies y t - 2 {t}^{2} = \left(x - {t}^{2}\right)$

$\implies x - y t + {t}^{2} = 0$

So length of the perpendicular drawn from $Q$ on the tangent drawn at $P$ will be

$= \frac{\frac{1}{t} ^ 2 - \left(- \frac{2}{t}\right) t + {t}^{2}}{\sqrt{{t}^{2} + 1}}$

$= {\left(t + \frac{1}{t}\right)}^{2} / \sqrt{{t}^{2} + 1}$

$= {\left({t}^{2} + 1\right)}^{2} / \left({t}^{2} \sqrt{{t}^{2} + 1}\right)$

By the condition of the problem

${\left({t}^{2} + 1\right)}^{2} / \left({t}^{2} \sqrt{{t}^{2} + 1}\right) = 2 \sqrt{2}$

$\implies {\left({t}^{2} + 1\right)}^{\frac{3}{2}} / {t}^{2} = 2 \sqrt{2}$

Let ${t}^{2} = a$

$\implies {\left(a + 1\right)}^{\frac{3}{2}} / a = 2 \sqrt{2}$

$\implies {\left(a + 1\right)}^{3} = {\left(2 \sqrt{2}\right)}^{2} {a}^{2}$

$\implies {a}^{3} + 3 {a}^{2} + 3 a + 1 - 8 {a}^{2} = 0$

$\implies {a}^{3} - 5 {a}^{2} + 3 a + 1 = 0$

$\implies {a}^{3} - {a}^{2} - 4 {a}^{2} + 4 a - a + 1 = 0$

$\implies {a}^{2} \left(a - 1\right) - 4 a \left(a - 1\right) - 1 \left(a - 1\right) = 0$

$\implies \left(a - 1\right) \left({a}^{2} - 4 a - 1\right) = 0$

When $a = 1$

$\implies {t}^{2} = 1$

$\implies t = \pm 1$

When

$\left({a}^{2} - 4 a - 1\right) = 0$

$\implies a = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot 1 \cdot \left(- 1\right)}}{2 \cdot 1}$
$\implies a = \frac{4 \pm \sqrt{20}}{2 \cdot 1}$

$\implies a = \frac{4 \pm 2 \sqrt{5}}{2 \cdot 1}$

$\implies {t}^{2} = \left(2 \pm \sqrt{5}\right)$

Neglecting imaginary roots we get

$t = \pm \sqrt{\sqrt{5} + 2}$

Hence the possible values of gradient $m = \frac{1}{t} = \pm 1 \mathmr{and} \pm \frac{1}{\sqrt{\sqrt{5} + 2}} = \pm \sqrt{\sqrt{5} - 2}$

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