You are Here: Home >< Maths

# Homework Help watch

1. The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
2. (Original post by Improvement)
The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
What'd dy/dx at x = pi/4? What does that tell you about the gradient of the normal? How can you then write the equation of the normal?
3. (Original post by Improvement)
The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
4. (Original post by Improvement)
The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
Differentiate and find the value for the gradient at that point. -1/m is th gradient of the normal. Construct an equation of the normal in the form y=mx+C. Then set mx+C equal to x to find Q.
5. (Original post by Zacken)
What'd dy/dx at x = pi/4? What does that tell you about the gradient of the normal? How can you then write the equation of the normal?
Yeah I have done that, 2tanx/cosx then subbed in pi/4 to get the gradient then put it -1/m to get the normals gradient. Subbed in pi/4 and 2 to get C; but I don't know how to find the co-ordinates of where x=y
6. (Original post by Improvement)
Yeah I have done that, 2tanx/cosx then subbed in pi/4 to get the gradient then put it -1/m to get the normals gradient. Subbed in pi/4 and 2 to get C; but I don't know how to find the co-ordinates of where x=y
So you have the equation of the line as y = something *x + c

You know that y=x

so x =something*x +c

x(1- something) = c

x = ... = y
7. (Original post by Improvement)
Yeah I have done that, 2tanx/cosx then subbed in pi/4 to get the gradient then put it -1/m to get the normals gradient. Subbed in pi/4 and 2 to get C; but I don't know how to find the co-ordinates of where x=y
Say the normal's equation is y=2x+1, to find the intersect with y=x, you set 2x+1=x which means x and y=-1. Do the same.
8. (Original post by Kvothe the arcane)
Say the normal's equation is y=2x+1, to find the intersect with y=x, you set 2x+1=x which means x and y=-1. Do the same.
Where did y=2x+1 come from?
9. (Original post by Improvement)
Where did y=2x+1 come from?
It was an arbitrary choice of linear equation. I can't solve your question for you but thought you might find it useful to be reminded (presumably you've done this before?) of the method.
10. (Original post by Kvothe the arcane)
It was an arbitrary choice of linear equation. I can't solve your question for you but thought you might find it useful to be reminded (presumably you've done this before?) of the method.
Yeah I have I just was unsure about where to go once I got to the equations y=mx+c

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 28, 2016
Today on TSR

### University open days

1. University of Cambridge
Wed, 26 Sep '18
2. Norwich University of the Arts
Fri, 28 Sep '18
3. Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 29 Sep '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams