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
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 $\displaystyle x^2-3x-4 = 0$. For checking purposed the co-ordinate of P is $\displaystyle (-1 , 0)$ and Q is $\displaystyle (4, 0)$. now you must find the equation of the tangents at P and Q. First differentiate $\displaystyle \frac{dy}{dx} = 2x - 3$. So the gradient of the tangent at P is $\displaystyle 2(-1) -3 = - 5$. to find the equation of the line use the formula $\displaystyle y - y_1 = m(x-x_1)$ so in this case $\displaystyle y - 0 = -5(x- (-1)) \ \ \rightarrow \ y= -5x - 5$. 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
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
Now you need to find the equation of the normals at P and Q. you know know that the gradient of the normal is $\displaystyle \frac{-1}{m}$ where m is the gradient of the tangent. so the gradient of the normal at P is $\displaystyle \frac{1}{5}$ and the equation of the normal is $\displaystyle y= \frac{1}{5}x + \frac{1}{5}$. Can you finish the question now ?
Bobak
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
To find the equation of the normals we use the formula $\displaystyle 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 $\displaystyle (-1,0)$ you have $\displaystyle y - 0 = \frac{1}{5}(x-(-1)) \ \rightarrow \ y= \frac{1}{5}x + \frac{1}{5}.$.
For Q the gradient of the normal is $\displaystyle -\frac{1}{5}$. can you now find the equation the normal at Q and finish off the question ?
Bobak
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?