Question

The slope of the normal to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (–10, –1) is 1/k. The value of k is

Answer 1

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

Given parametric curve is

$\rm :\longmapsto\:x = {t}^{2} + 3t - 8$

and

$\rm :\longmapsto\:y = {2t}^{2} - 2t - 5$

Since, it is given that, Point of contact is ( - 10, - 1 ).

So,

$\rm :\longmapsto\: - 10 = {t}^{2} + 3t - 8$

$\rm :\longmapsto\: {t}^{2} + 3t - 8 + 10 = 0$

$\rm :\longmapsto\: {t}^{2} + 3t + 2 = 0$

$\rm :\longmapsto\: {t}^{2} + 2t + t + 2 = 0$

$\rm :\longmapsto\:t(t + 2) + 1(t + 2) = 0$

$\rm :\longmapsto\:(t + 2)(t + 1) = 0$

$\bf\implies \:t = - 1 \: \: or \: \: t = - 2$

Also,

$\rm :\longmapsto\: - 1 = {2t}^{2} - 2t - 5$

$\rm :\longmapsto\: {2t}^{2} - 2t - 5 + 1 = 0$

$\rm :\longmapsto\: {2t}^{2} - 2t -4 = 0$

$\rm :\longmapsto\: {t}^{2} - t -2 = 0$

$\rm :\longmapsto\: {t}^{2} - 2t + t -2 = 0$

$\rm :\longmapsto\:t(t - 2) + 1(t - 2) = 0$

$\rm :\longmapsto\:(t + 1)(t - 2) = 0$

$\bf\implies \:t = - 1 \: \: or \: \: t = 2$

So,

It means we have to find the slope of normal at t = - 1.

Now,

$\rm :\longmapsto\:x = {t}^{2} + 3t - 8$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt} x = \dfrac{d}{dt}( {t}^{2} + 3t - 8)$

$\rm :\longmapsto\:\dfrac{dx}{dt} = 2t + 3$

Also,

$\rm :\longmapsto\:y = {2t}^{2} - 2t - 5$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}y =\dfrac{d}{dt}( {2t}^{2} - 2t - 5)$

$\rm :\longmapsto\:\dfrac{dy}{dt} =4t - 2$

So,

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{4t - 2}{2t + 3}$

Thus, Slope of tangent at t = - 1, is given by

$\rm :\longmapsto\:\boxed{ \tt{ \: Slope \: of \: tangent, \: m = \bigg[\dfrac{dy}{dx}\bigg] _{t = - 1}}}$

So,

$\rm :\longmapsto\: Slope \: of \: tangent, \: m = \dfrac{4( - 1) - 2}{2( - 1) + 3}$

$\rm :\longmapsto\: Slope \: of \: tangent, \: m = \dfrac{ - 4 - 2}{ - 2 + 3}$

$\rm :\longmapsto\: Slope \: of \: tangent, \: m = \dfrac{ - 6}{1}$

$\bf\implies \: Slope \: of \: tangent, \: m = - 6$

Now, we know,

$\boxed{ \tt{ \: Slope \: of \: normal = - \: \frac{1}{Slope \: of \: tangent} \: }}$

$\bf\implies \: \: Slope \: of \: normal = - \: \dfrac{1}{ - 6} = \dfrac{1}{6}$

As it is given that,

$\rm :\longmapsto\:Slope \: of \: normal \: = \: \dfrac{1}{k}$

So, On comparing we get,

$\bf \implies\:\boxed{ \bf{ \: k \: = \: 6 \: }}$

Additional Information :-

1.bLet y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined.

4. Two lines having slope M and m are parallel, iff M = m

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.