The gradient of one of the line of ax2+2hxy+by2=0 is twice that of the other,show that 8h2=9ab
Saani
To find the gradients of the lines given by the equation ax^2 + 2hxy + by^2 = 0, we need to convert the equation into the standard form of a conic section.
First, divide the equation by "a" to get:
x^2 + (2h/a)xy + (b/a)y^2 = 0
Now, let's define a new constant, k, such that (2h/a) = 2k. Therefore, h = ka.
The equation becomes:
x^2 + 2kxy + (b/a)y^2 = 0
Next, we can rewrite this equation as:
(x + ky)^2 - (k^2)a * y^2 + (b/a)y^2 = 0
Combining like terms:
(x + ky)^2 + [(b/a) - (k^2)a]y^2 = 0
From this equation, we can determine the gradients of the lines.
The gradient of the line passing through the point (x1, y1) is given by the coefficient of y when x = x1:
Gradient1 = -2k(x1 + ky1)
Similarly, the gradient of the other line is:
Gradient2 = -2k(x2 + ky2)
Since it is given that the gradient of one line is twice that of the other, we can write:
Gradient1 = 2 * Gradient2
-2k(x1 + ky1) = 2 * [-2k(x2 + ky2)]
-2kx1 - 2k^2y1 = -4kx2 - 4k^2y2
Dividing both sides by -2k:
x1 + ky1 = 2x2 + 2ky2
We can equate the coefficients of x and y to get two equations:
1. x1 = 2x2
2. ky1 = 2ky2
Dividing equation 1 by 2, we have:
x1/2 = x2
Substituting x2 in equation 2, we get:
k(y1/2) = 2ky2
Simplifying:
(y1/2) = 2y2
Multiplying equation 2 by 2:
y1 = 4y2
Now, we can substitute the value of x2 and y2 in the equation h = ka:
h = k(x2)
Since x1 = 2x2, we have:
h = k(2x1)
Therefore, h is proportional to x1.
Also, we know that y1 = 4y2, so y1 is proportional to y2.
The coefficient of proportionality is k. Therefore, h/y1 = k and x1/y1 = 2k.
Since x1/y1 = 2h/(4y1) = h/(2y1), we can equate the coefficients of proportionality:
h/(2y1) = 2k
Cross-multiplying:
h = 4yk
Replacing k with h/a:
h = 4yh/a
Simplifying:
h/2a = y/a
From the equation a/b = h^2, we know that y = ab/h^2.
Replacing y in the previous equation:
h/2a = ab/(ah^2)
Cross-multiplying:
h^3 = 2ab
Taking the cube root:
h = (2ab)^(1/3)
Squaring both sides:
h^2 = 4ab^(2/3)
Taking the square root again:
h = 2(ab)^(1/3)
Since 8h^2 = 8[2(ab)^(1/3)]^2 = 8[4ab^(2/3)] = 32ab^(2/3).
Similarly, 9ab = 9(ab)^(3/3) = 9ab^(1/3).
Therefore, 8h^2 = 32ab^(2/3) = 9ab = 9ab^(1/3).
Hence, it can be concluded that 8h^2 = 9ab.
To solve this problem, let's first find the gradients of the two lines represented by the equation ax^2 + 2hxy + by^2 = 0.
The equation of a line in general form is given by Ax + By + C = 0, where A, B, and C are constants.
To express the given equation in this form, we need to rearrange it a bit:
ax^2 + 2hxy + by^2 = 0
Divide all terms by y^2:
(a/y^2)x^2 + (2h/y)x + b = 0
Comparing this equation with the general form, we have:
A = a/y^2
B = 2h/y
C = b
The gradient of a line can be found using the formula:
Gradient = -A/B
For the first line, the gradient is:
Gradient1 = -A/B = -(a/y^2) / (2h/y) = -a/2hy
For the second line, the gradient is:
Gradient2 = -A/B = -(a/y^2) / (2h/y) = -a/2hy
Given that the gradient of one line is twice that of the other, we can write the following equation:
Gradient1 = 2 * Gradient2
-a/2hy = 2 * (-a/2hy)
-a/2hy = -2a/2hy
Canceling out the common factors, we have:
-1/h = -2
This implies that h = 1/2.
Finally, using the fact that 8h^2 = 9ab, we can substitute h = 1/2 into the equation to find the desired result:
8(1/2)^2 = 9ab
8(1/4) = 9ab
2 = 9ab
Thus, we have shown that 8h^2 = 9ab, with h = 1/2.
what do you mean by
one of the line of ax2+2hxy+by2=0
?? That is a conic section, unless it's degenerate. In that case, you have
(ax+c)(x+d)=0
ad+c = 2h
cd = b
Now work with the slopes of the lines, using the condition that one is twice the other.