if the line x+y+k is a normal to the hyperbola xsquare/9-ysquare/4=1 then k=
I will assume you meant:
if the line x+y = k is a normal to the hyperbola x^2/9-y^2/4 = 1 then k=
your hyperbola is 4x^2 - 9y^2 = 36
8x - 18y dy/dx = 0
dy/dx = 8x/18y = 4x/9y
at any point on the hyperbola, the slope of a tangent is 4x/9y
so the slope of the corresponding normal would be -9y/4x
but the slope of x + y = k is -1
so -9y/4x = -1
9y = 4x
y = 4x/9
sub into 4x^2 - 9y^2 = 36
4x^2 - 9(16x^2/81) = 36
4x^2 - 16x^2/9 = 36
36x^2 - 16x^2 = 324
x^2 = 324/20 = 81/5
x = ± 9/√5
then y = ± 4x/9 = ±4(9/√5)/9 = ± 4/√5
but from my sketch , x+y = k can only be in quadrants I or III
using (9/√5 , 4/√5)
9/√5 + 4/√5 = k = 13/√5
x + y = 13/√5 in quad I
x + y = -13/√5 in quad III
checking with Wolfram:
www.wolframalpha.com/input/?i=solve+4x%5E2+-+9y%5E2+%3D+36+,+x+%2B+y+%3D+13%2F%E2%88%9A5
Wolfram has both graphed and solved my equations.
To determine the value of k for which the line x + y + k is normal to the hyperbola, we need to use the concept of slopes.
The equation of the hyperbola is given as xsquare/9 - ysquare/4 = 1. To find the slope of the hyperbola at a given point, we need to differentiate the equation of the hyperbola with respect to x and solve for dy/dx.
Differentiating the given equation, we get:
2x/9 - 2y(dy/dx)/4 = 0
Simplifying the equation gives us:
2x/9 - y(dy/dx)/2 = 0
Multiplying through by 2/9, we have:
x - y(dy/dx)(2/9) = 0
Now, the slope of the line x + y + k is given by the coefficient of x in its equation, which is 1.
Since a line is normal to a curve when the product of their slopes is -1, we can equate the product of the slopes to -1 and solve for k.
1 * (-y(dy/dx)(2/9)) = -1
Simplifying the equation gives us:
-y(dy/dx)(2/9) = -1
Dividing both sides by -y(dy/dx)(2/9), we have:
-1 / (-y(dy/dx)(2/9)) = -1 / (-y(dy/dx)(2/9))
Simplifying further:
1 / (y(dy/dx)(2/9)) = 1 / 1
Canceling out the 1's on both sides, we get:
y(dy/dx)(2/9) = 1
Substituting the value of dy/dx from the derivative equation we derived earlier, we have:
y[(2x/9) * (2/9)] = 1
Simplifying the equation gives us:
y(4x/81) = 1
Dividing both sides by (4x/81), we have:
y = 81 / (4x)
Now we need to find the value of k. To do that, substitute the value of y from the above equation into the equation of the line x + y + k = 0:
x + (81 / (4x)) + k = 0
Multiply through by 4x to eliminate the denominator:
4x^2 + 81 + 4kx = 0
This equation represents a quadratic equation in x. For the line to be normal to the hyperbola, this quadratic equation should have exactly one root. In other words, its discriminant should equal zero.
The discriminant of the quadratic equation is given by b^2 - 4ac. In this case, a = 4, b = 4k, and c = 81.
Setting the discriminant equal to zero, we have:
(4k)^2 - 4(4)(81) = 0
Simplifying the equation gives us:
16k^2 - 4(4)(81) = 0
Now, solve for k by rearranging the equation and dividing both sides by 16:
16k^2 = 4(4)(81)
Dividing both sides by 4 gives us:
4k^2 = 324
Dividing both sides by 4 again gives us:
k^2 = 81
Taking the square root of both sides, we have:
k = ± 9
Therefore, the value of k such that the line x + y + k is normal to the hyperbola xsquare/9 - ysquare/4 = 1 is k = ± 9.