Login

Welcome, Guest. Please login or register.

April 19, 2024, 03:28:00 pm

Author Topic: I am clueless about this question, Can someone please help  (Read 583 times)  Share 

0 Members and 1 Guest are viewing this topic.

Rose34

  • Trendsetter
  • **
  • Posts: 158
  • Respect: +2
I am clueless about this question, Can someone please help
« on: February 27, 2020, 10:49:35 pm »
+1
Find the equation of the straight line(s) which pass through the point (1,−2) and is (are) tangent to the parabola with equation y = x^2.

I really want to know how to solve this question so if anyone have any idea please help.

Thank you

BiggestVCESweat

  • Guest
Re: I am clueless about this question, Can someone please help
« Reply #1 on: February 29, 2020, 11:24:34 am »
+3
Method 1: (Techfree)
If it is tangent to the parabola, that means that when you make the equations intersect the discriminant will be 0.
The equation of the line is y=mx+c
hence if you let them equal each other:
y=x^2 - mx - c
Discriminant = m^2 - 4c = 0
Since the line passes through (1, -2)
-2 = m + c

So then you have two equations, m^2 - 4c=0, -2 = m+c, and you just solve for m and c and you get your tangent lines.


Method 2 (cas):
solve((tangentLine(x^2,x,m)|x=1)=-2,m)
Then just plug the values of m you get back into tangentLine.

What the command above does is it solves for the point the tangent line is tangent to on the curve, given that when x=1 the tangent line gives -2. Then when you plug the values of m back into the function it gives you the expression of the tangentLine. (You have to add y= before it.)


You can also use calculus, but I will leave that as an exercise for the reader.

Rose34

  • Trendsetter
  • **
  • Posts: 158
  • Respect: +2
Re: I am clueless about this question, Can someone please help
« Reply #2 on: February 29, 2020, 08:42:57 pm »
+1
Method 1: (Techfree)
If it is tangent to the parabola, that means that when you make the equations intersect the discriminant will be 0.
The equation of the line is y=mx+c
hence if you let them equal each other:
y=x^2 - mx - c
Discriminant = m^2 - 4c = 0
Since the line passes through (1, -2)
-2 = m + c

So then you have two equations, m^2 - 4c=0, -2 = m+c, and you just solve for m and c and you get your tangent lines.


Method 2 (cas):
solve((tangentLine(x^2,x,m)|x=1)=-2,m)
Then just plug the values of m you get back into tangentLine.

What the command above does is it solves for the point the tangent line is tangent to on the curve, given that when x=1 the tangent line gives -2. Then when you plug the values of m back into the function it gives you the expression of the tangentLine. (You have to add y= before it.)


You can also use calculus, but I will leave that as an exercise for the reader.

Thank you so much