Question

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Answer 1

Step-by-step explanation:

★ Concept :-

Here the concept of conic sections and differentiation has been used. We see that we are given equation of a curve where the tangent lines are parallel to another line. So this means the slope of both the lines will be same. So firstly we can find the slopes of both the lines. Then we can equate them. After equating we will get the points.

Let's do it !!

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★ Solution :-

Given,

• » Equation of curve : y = x³ - 3x² - 4x
• » Another line : 4x + y - 3 = 0

Let the point on curve be (h, k)

This means h is on the curve so we can equate this h in slope of the equation.

Since these lines (tangent line and another line) are parallel, so their slope will be same.

Let the slope of the lines be m

• • For the slope of tangent line on curve ::

We can find this using differentiation. Then,

$\;\tt{\rightarrow\;\;y'\;=\;\dfrac{dy}{dx}\;=\;\dfrac{d}{dx}(x^{3}\:-\:3x^{2}\:-\:4x)}$

We know that derivates are distributive in nature. So,

$\;\tt{\rightarrow\;\;y'\;=\;\dfrac{d}{dx}(x^{3})\:-\:\dfrac{d}{dx}(3x^{2})\:-\:\dfrac{d}{dx}(4x^{1})}$

$</p><p>\;\tt{\rightarrow\;\;y'\;=\;(3x^{3\:-\:1})\:-\:(3(2x^{2\:-\:1}))\:-\;(4(1x^{1\:-\:1}))}$

$\;\tt{\rightarrow\;\;y'\;=\;(3x^{3\:-\:1})\:-\:(3(2x^{2\:-\:1}))\:-\:(4(1x^{1\:-\:1}))}$

$\;\tt{\rightarrow\;\;y'\;=\;(3x^{2})\:-\:(3(2x^{1}))\:-\:(4(1x^{0}))}$

$\;\tt{\rightarrow\;\;y'\;=\;(3x^{2})\:-\:(3(2x))\:-\:(4(1(1)))}$

$\;\tt{\rightarrow\;\;y'\;=\;(3x^{2})\:-\:(3(2x))\:-\:(4(1(1)))}$

$\;\bf{\rightarrow\;\;y'\;=\;3x^{2}\:-\:6x\:-\:4}$

Now here let's equate the point h in place of x here. Then,

>> y' = 3h² - 6h - 4

• For the slope of another line ::

We know that slope of the any line is given as,

→ y = mx + c (where m is the slope)

So we have the equation of line as,

>> 4x + y - 3 = 0

>> y = -4x + 3

On comparing this with the equation of slope, we get

>> m = -4 , c = 3

• • For the value of h ::

We already got the slopes of the parallel lines. Since they are equal, so

>> 3h² - 6h - 4 = -4

Cancelling -4 from both sides, we get

>> 3h² - 6h = 0

>> 3h(h - 2) = 0

Here either 3h = 0 or (h - 2) = 0

So,

>> 3h = 0 or h - 2 = 0

>> h = 0/3 or h = 2

>> h = 0 or h = 2

• • For the coordinates of point :

We see that we equated the value of h in the derivative of the curve. The we can write initial equation as,

>> k = h³ - 3h² - 4h

(on comparing it with x and y coordinates of of curve)

• when h = 0 ::

>> k = 0³ - 3(0)² - 4(0)

>> k = 0 - 0 - 0

>> k = 0

Hence the coordinate will be (0, 0)

• when h = 2

>> k = 2³ - 3(2)² - 4(2)

>> k = 8 - 3(4) - 8

>> k = -12

Hence, the coordinate will be (2, -12)

This gives the points on the curve.

$\;\bigg|\underline{\boxed{\tt{Points\;\:on\;\:the\;\:curve\;=\;\bf{\purple{(0,0)}}\;\tt{or}\;\bf{\purple{(2, -12)}}}}}\bigg|$