find the lines that are (a) tangent and (b) normal to the curve at the given point
x^2 + xy - y^2 = 1
Take the differential...
2xdx+ y dx + x dy -2ydy=0
solve for dy/dx
Then use that as m in
y=mx + b Puttingin the x,y point, solve for b.
then, for the normal, take the negative reciprocal of m, and again, solve for the line
y=-1/m x + b
You did not state the given point.
Using implicit derivative I found it to be
y' = (2x+y)/(2y-x)
sub in the given point, that gives you the slope of the tangent.
Now that you have the slope (m) and a given point, use the grade 9 method of finding the equation of the tangent.
Take the negative reciprocal of your slope, and the given point to find the equation of the normal.
You forgot to say what point.
However in general we find the derivative, dy/dx and call that the slope, m, of the tangent at the point. Then plug the x and y of the point in to get b in y = m x + b
to get the derivative
2 x dx/dx +x dy/dx + y dx/dx - 2 y dy/dx = 0
or
2x + y = (2y-x) dy/dx
dy/dx = (2x+y)/(2y-x)
Well, well, well, let's find those tangents and normals, shall we? Strap on your mathematical seatbelt and away we go!
To find the lines that are tangent and normal to the curve, we'll first need to find the derivative of the equation. Taking the derivative of x^2 + xy - y^2 = 1 with respect to x, we get:
2x + y + x(dy/dx) - 2y(dy/dx) = 0
Now, let's find the slope of the tangent line at a given point. We'll substitute the values of x and y into the newly derived equation and solve for dy/dx. Let's choose a point like (3, 2) just for fun:
2(3) + 2 + 3(dy/dx) - 2(2)(dy/dx) = 0
Simplifying this equation, we get:
6 + 2 + 3(dy/dx) - 4(dy/dx) = 0
8 - dy/dx = 0
Now, we solve for dy/dx:
dy/dx = 8
So, the slope of the tangent line at the point (3, 2) is 8.
To find the equation of the tangent line, we'll use the point-slope form of a line: y - y1 = m(x - x1). Plugging in the values, we get:
y - 2 = 8(x - 3)
Simplifying this equation, the equation of the tangent line is:
y = 8x - 22
As for the normal line, since the slope of the tangent line is 8, the slope of the normal line will be the negative reciprocal of 8, which is -1/8. We'll use the same point-slope form, but this time with the slope of -1/8:
y - 2 = (-1/8)(x - 3)
Simplifying this equation, the equation of the normal line is:
8y + x = 26
Ta-da! We have found the tangent line, y = 8x - 22, and the normal line, 8y + x = 26, to the curve at the point (3, 2).
To find the lines that are tangent and normal to the curve at a given point, we need to find the derivative of the curve, since the derivative represents the slope of the tangent line at any given point on the curve.
Let's find the derivative of the curve x^2 + xy - y^2 = 1 using implicit differentiation:
Step 1: Differentiate both sides of the equation with respect to x.
d/dx (x^2 + xy - y^2) = d/dx (1)
Step 2: Apply the product and chain rules to differentiate each term separately.
2x + x(d/dx)(y) + y - 2y(dy/dx) = 0
Step 3: Simplify the equation.
2x + xy' + y - 2yy' = 0
Step 4: Rearrange the equation to isolate y'.
xy' - 2yy' = -2x - y
y'(x - 2y) = -2x - y
Step 5: Solve for y'.
y' = (-2x - y)/(x - 2y)
Now that we have obtained an expression for y', we can find the slope of the tangent line at any given point on the curve by plugging the coordinates of the point into the derivative expression.
Let's assume we have a given point (a, b) on the curve. Plugging these values into the derivative expression, we obtain:
y' = (-2a - b)/(a - 2b)
(a) Tangent Line: The equation of the tangent line passing through the given point (a, b) will have a slope equal to y'. We can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:
y - b = y'(x - a)
Substituting the slope y' = (-2a - b)/(a - 2b) and the point coordinates (a, b) into the point-slope form, we get:
y - b = [(-2a - b)/(a - 2b)](x - a)
Simplifying this equation will give us the equation of the tangent line.
(b) Normal Line: The normal line is perpendicular to the tangent line at the given point. Since perpendicular lines have negative reciprocal slopes, the slope of the normal line will be the negative reciprocal of y'. We can find the equation of the normal line using the point-slope form similar to the tangent line.
The equation of the normal line passing through the given point (a, b) will be:
y - b = -[1/y'(x - a)]
Substitute y' = (-2a - b)/(a - 2b) and the point coordinates (a, b) into the equation to obtain the equation of the normal line.
Note: To find the specific lines tangent and normal to the curve at a particular point, you need to substitute the values of (a, b) into the equations obtained in steps (a) and (b).
thank alot
4) x - 4y = -31
2x - 4y = -34