Q32. Find the points on the curve y = x³ - 3x² - 4x at which the tangent lines are parallel to the line 4x + y - 3 = 0
Answers
★ 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,
We know that derivates are distributive in nature. So,
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.
Answer:
At point (0 ,0) and (2 , -12), tangent to curve y = x³ - 3x² - 4x is parallel to line 4x + y - 3 = 0 .
Step-by-step explanation:
- A tangent is a straight line which touches the curve y = f(x) at a point.
- The derivative of the curve y = f(x) is f'(x) which represents the slope of tangent and equation of the tangent to the curve at P is (Y - y) = f'(x)(X - x), where (x , y) is an arbitrary point on the tangent.
- Slope of line y = mx +c is m.
Given that :
- Curve, y = x³ - 3x² - 4x
- line, L : 4x + y - 3 = 0
To find :
- Point on curve at which tangent lines is parallel to line L.
Solution :
- The given equation of curve is y = x³ - 3x² - 4x
- Slope of the curve is
- f'(x) = 3x² - 6x - 4 ---(1)
- Let, (h , k) be the point on curve such that tangent at (h , k) be parallel to line L.
- Slope of the curve at point (h , k) is
- (3h² - 6h - 4)
- Slope of line L is y = -4x + 3. Hence, slope of line L is -4.
- The slope of curve at (h , k) is equal to slope of line L . This gives,
⇒3h² - 6h - 4 = -4
⇒3h² - 6h = 0
⇒3h(h - 2) = 0
⇒h = 0 or h = 2.
- Putting h = 0 in curve's equation gives k = 0.
- Putting h = 2 in curve's equation gives k = -12.
- Hence, at point (0 ,0) and (2 , -12), tangent to curve is parallel to line L.
- Equation of tangent at point (0 , 0) is
⇒(y - 0) = -4(x - 0)
⇒y = - 4x ---(2)
- Equation of tangent at point (2 , -12) is
⇒(y - (-12)) = -4(x - 2)
⇒4x + y + 4 = 0 ---(3)