Question

#Quality Question
@Tangent and Normals

Find all points on the curve
$y = 4 {x}^{3} - 2 {x}^{5}$
at which the tangent passes through origin​

Answer 1

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

Given curve is

$\rm :\longmapsto\:y = {4x}^{3} - {2x}^{5} - - - (1)$

Let assume that the point of contact of tangent with curve be P (h, k).

As, P lies on the curve (1), so

$\rm :\longmapsto\:k = {4h}^{3} - {2h}^{5} - - - (2)$

Since, Tangent passes through origin.

We know that,

Slope of a line joining two points ( a, b ) and ( c, d ) is represented by 'm' and given by

$\red{\boxed{ \bf{ \: m \: = \: \frac{d - b}{c - a}}}}$

Now, we have

Coordinates of O (0, 0)

and

Coordinates of P ( h, k ).

So,

$\rm :\longmapsto\:Slope \: of \: tangent, \: OP\: = \: \dfrac{k - 0}{h - 0} = \dfrac{k}{h} - - (3)$

Now, given curve

$\rm :\longmapsto\:y = {4x}^{3} - {2x}^{5}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx}( {4x}^{3} - {2x}^{5} )$

$\rm :\longmapsto\:\dfrac{dy}{dx} =4 \dfrac{d}{dx}{x}^{3} -2\dfrac{d}{dx} {x}^{5}$

$\rm :\longmapsto\:\dfrac{dy}{dx} =4 ( {3x}^{2}) -2( {5x}^{4})$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {12x}^{2} -{10x}^{4}$

Therefore, slope of tangent OP at P ( h, k ) is

$\rm :\longmapsto\:Slope \: of \: OP \: = \dfrac{dy}{dx}_{( h, k )} = {12h}^{2} -{10h}^{4} - - - (4)$

Now, Equating equation (3) and (4), we get

$\rm :\longmapsto\:\dfrac{k}{h} = {12h}^{2} - {10h}^{4}$

$\rm :\longmapsto\:k = {12h}^{3} - {10h}^{5}$

On substituting the value of k from equation (2), we get

$\rm :\longmapsto\: {4h}^{3} - {2h}^{5} = {12h}^{3} - {10h}^{5}$

$\rm :\longmapsto\:{8h}^{3} - {8h}^{5} = 0$

$\rm :\longmapsto\:{8h}^{3}(1 - {h}^{2}) = 0$

$\bf\implies \:h = 0 \: \: or \: h \: = \: \pm \: 1$

So, value of k to the corresponding value of h is

$\begin{gathered}\boxed{\begin{array}{c|c} \bf h & \bf k = {4h}^{3} - {2h}^{5} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf 2 \\ \\ \sf - 1 & \sf - 2 \end{array}} \\ \end{gathered}$

Hence,

The points on the curve at which tangent passes through origin are ( 0, 0 ), ( 1, 2 ) and ( - 1, - 2 )

Additional Information :-

1. Let 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.

Answer 2

*Question:-

Find all points on the curve y= 4x³ - 2x⁵ at which the tangent passes through origin

*Answer:-

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