# Thread: tangents and normals to curve

1. ## tangents and normals to curve

the curve y=x^2-3x-4 crosses the x axis a P and Q. The tangets to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS?

Really struggling thanks fo any help

2. Originally Posted by Soloman
the curve y=x^2-3x-4 crosses the y axis a P and Q. The tangets to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS?

Really struggling thanks fo any help
the graph given crosses the y-axis once. please review your question and make sure of the wording

3. sorry meant to be x axis sorry

4. Originally Posted by Soloman
the curve $y=x^2-3x-4$ crosses the x axis a P and Q. The tangets to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS?

Really struggling thanks fo any help

First you find x co-ordinate of P and Q by solving the equation $x^2-3x-4 = 0$. For checking purposed the co-ordinate of P is $(-1 , 0)$ and Q is $(4, 0)$. now you must find the equation of the tangents at P and Q. First differentiate $\frac{dy}{dx} = 2x - 3$. So the gradient of the tangent at P is $2(-1) -3 = - 5$. to find the equation of the line use the formula $y - y_1 = m(x-x_1)$ so in this case $y - 0 = -5(x- (-1)) \ \ \rightarrow \ y= -5x - 5$. I'll leave finding the equation of the tangent at Q up to you.

Now you must find the intersection point of the tangents (that is R) and then the intersection points of the normals (that is S). then find the distance between.

Bobak

5. Originally Posted by Soloman
sorry meant to be x axis sorry
well, then, we can find P and Q by setting y = 0 and solving for x, right.

once we have P and Q, we can find the tangent lines at P and Q, these are just lines with slope given by the derivative at these points passing through P and Q respectively. the normal lines are the lines perpendicular to those through the same points.

once you have all the lines, you can find their intersections to get R and S. then the distance between R and S using the distance formula

6. So I found the equations for P and Q they are:
P y=-5x-5
Q y=5x-20

then I found the point of intersection then what do I do?

x=1.5 and y=-12.5

7. Originally Posted by Soloman
So I found the equations for P and Q they are:
P y=-5x-5
Q y=5x-20

then I found the point of intersection then what do I do?

x=1.5 and y=-12.5
Now you need to find the equation of the normals at P and Q. you know know that the gradient of the normal is $\frac{-1}{m}$ where m is the gradient of the tangent. so the gradient of the normal at P is $\frac{1}{5}$ and the equation of the normal is $y= \frac{1}{5}x + \frac{1}{5}$. Can you finish the question now ?

Bobak

8. why is it 1/5 for c?

9. Can someone please show me how to figure it out ive been trying for ages

10. Originally Posted by Soloman
Can someone please show me how to figure it out ive been trying for ages
if two lines are perpendicular (in particular, the normal line and the tangent line are perpendicular) then the product of their slopes is -1. that means, their slopes are the negative inverses of each other

you had the slope of one of the tangent lines to be -5, thus, the slope of the corresponding normal line is the negative inverse of that, namely 1/5

11. thanks but what bout the rest

12. Originally Posted by Soloman
why is it 1/5 for c?
To find the equation of the normals we use the formula $y - y_1 = m(x-x_1)$ as we did before for the tangents. however the gradients of the normal is the negative reciprocal of the gradient of the tangents.

For example, the gradient of the tangent at P is -5 so the gradient of the normal is 1/5. then using the equation with the point $(-1,0)$ you have $y - 0 = \frac{1}{5}(x-(-1)) \ \rightarrow \ y= \frac{1}{5}x + \frac{1}{5}.$.

For Q the gradient of the normal is $-\frac{1}{5}$. can you now find the equation the normal at Q and finish off the question ?

Bobak

13. okay so I have
p = y=1/5x+1/5
q= -1/5x+4/5

so now I try and find the point of intersection right by making them equal each other then that will give me the x then I put it into one of the first equations to give me y. Then I have found coordinate of R. So how do you make an equation of a line that is perpendicular to R when u only have its coordinates?

14. Originally Posted by Soloman
So how do you make an equation of a line that is perpendicular to R when u only have its coordinates?
I don't understand why you are asking this, your question is to find the distance between S and R. do you know how to find the distance between two points ?

Bobak

15. okay so this is what I have for R its coordinates are x=1.5 and y=0.5
S is the normal of R is that right?

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