(A) Find the equations of the tangent lines to the circle x^2 +y^2=25 at the points where x=4.
(B) Find the equations of the normal lines to this circle at the same points. (The normal line is perpendicular to the tangent line at that point.)
(C) At what point do the two normal lines intersect?
For part A, I found that
16 + y 2= 25
y 2= 9
y=3
y=3/4x
For part B,I'm not sure whether it is y=-4/3x
For part C Not sure how to do it.
I solved the question in A but please check..
first of all you have to find y while x=4 and as you find y^2=25-16=9 and y=3
then you have to find slope. It's derivative of y function so
(slope)m=1/2(1/25-x^2)(2x).
if we write 3 instead of x we will get m(slope) as -4/3.
from
the tangent line equation
y-3=(-4/3)(x-4)
y=(-4/3)x+25/3.(result)
solution for part B
for the normal line equation firstly you have to find again slope. product of tangentline's slope and normal line's slope must be -1.
there fore
line's slope.(-4/3)=-1
line's slope= 3/4
then
the line equation is
y-3= =(3/4)(x-4)
we get
y= 3x/4(result)
For part A, to find the equation of the tangent lines to the circle at the points where x=4, we need to find the corresponding y-values.
Given that x=4, we can substitute this value into the equation of the circle:
x^2 + y^2 = 25
4^2 + y^2 = 25
16 + y^2 = 25
y^2 = 9
y = ±3
So the two points on the circle where x=4 are (4, 3) and (4, -3).
To find the equations of the tangent lines, we need to find the slopes of the tangents at these points. The slope of a tangent line to a circle at a given point is equal to the negative reciprocal of the slope of the radius passing through that point.
The equation of the circle x^2 + y^2 = 25 can be expressed as y = ±√(25 - x^2).
Taking the derivative of y with respect to x, we get:
dy/dx = ±(-x/√(25 - x^2))
Now, substituting x=4 into the derivative equation:
dy/dx = ±(-4/√(25 - 4^2))
= ±(-4/√(25 - 16))
= ±(-4/√9)
= ±(-4/3)
So, the slopes of the tangent lines are ±(-4/3).
Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can write the equations of the tangent lines:
1. For the point (4, 3) with slope -4/3:
y - 3 = -(4/3)(x - 4)
Simplifying, we get:
y - 3 = -(4/3)x + 16/3
y = -(4/3)x + 25/3
2. For the point (4, -3) with slope 4/3:
y + 3 = (4/3)(x - 4)
Simplifying, we get:
y + 3 = (4/3)x - 16/3
y = (4/3)x - 25/3
So, the equations of the tangent lines to the circle x^2 + y^2 = 25 at the points where x=4 are y = -(4/3)x + 25/3 and y = (4/3)x - 25/3.
For part B, to find the equations of the normal lines to the circle at the same points, we need to find the slopes of the normal lines. The slope of a normal line is the negative reciprocal of the slope of the tangent line at that point.
For the tangent line with slope -4/3, the normal line's slope will be the negative reciprocal, which is 3/4.
For the tangent line with slope 4/3, the normal line's slope will be the negative reciprocal, which is -3/4.
Using the point-slope form again, we can write the equations of the normal lines:
1. For the point (4, 3) with slope 3/4:
y - 3 = (3/4)(x - 4)
Simplifying, we get:
y - 3 = (3/4)x - 3
y = (3/4)x
2. For the point (4, -3) with slope -3/4:
y + 3 = (-3/4)(x - 4)
Simplifying, we get:
y + 3 = (-3/4)x + 3
y = (-3/4)x
So, the equations of the normal lines to the circle x^2 + y^2 = 25 at the points where x=4 are y = (3/4)x and y = (-3/4)x.
For part C, to find the point of intersection of the two normal lines, we need to solve the system of equations formed by setting the two normal lines equal to each other.
Setting y = (3/4)x and y = (-3/4)x, we can equate these two equations:
(3/4)x = (-3/4)x
Simplifying, we get:
6/4 x = 0
Dividing both sides by 6/4, we get:
x = 0
Substituting this value of x back into one of the equations, let's use y = (3/4)x:
y = (3/4)(0)
y = 0
Therefore, the two normal lines intersect at the point (0, 0).