How to find the coordinates of a point parallel to a tangent?
Answers
Step-by-step explanation:
Determine the slope of the given line: Since we want the tangent line to be parallel to the line 2x−4y=8, we need for it to have the same slope. We can rewrite the equation in slope-intercept form to easily find slope: 4y=2x−8⟺y=12x−2. Hence the desired slope of our line is m=12.
Then find the derivative of y=1−x2 and set it equal to the desired slope. y′=−2x. So put −2x=12
Solve for the value of x.
−2x=12⟺x=−14
That will give you the x-coordinate of a point on the desired line. Find the corresponding y-coordinate, and you'll have both coordinates of a point on the desired line.
y=1−x2⟹y(−14)=1−116=1516
We know the slope of the desired line is equal to 12, and we now have a point on the desired line, as well as on the given curve: i.e., the point of tangency (−14,1516).
Use the point-slope form to obtain the equation of the desired line. Given slope m and a point on the desired line,(x0,y0), the equation of the line is given by
(y−y0)=m(x−x0)
This gives us:
y−1516=12(x+14)⟺16y−15=8x+2⟺8x−16y=−17
Answer:
if u r line bisect we can say it is parallel to a tangent