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The gradient of the curve y px qy=0 at (1,1) is1/2.the values of P and q are?
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The gradient of the curve y px qy=0 at (1,1) is1/2.the values of P and...
y+px+qy=0

y+qy=-px

(1+q)y=-px

y=-[p/(1+q)]x

This is a linear function, so has constant gradient at all points on the curve. Hence

-p/(1+q)=1/2

2p=-(1+q)

But there is an issue: you have stated that the curve's gradient is 1/2 at the point (1,1) but the curve does not cross through this point! Regardless of our choices for p and q satisfying the expressions above this paragraph, the equation of the curve will always simplify to y=0.5x, which crosses through the origin (0,0), as well as (1,0.5) and (2,1) - but not (1,1).

For your curve to pass through (1,1), we would need to add a constant term, like so:

y+px+qy=1/2
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The gradient of the curve y px qy=0 at (1,1) is1/2.the values of P and...
Given information:
The equation of the curve is y px qy = 0.
The gradient of the curve at the point (1, 1) is 1/2.

Explanation:
To find the values of p and q, we need to differentiate the given equation with respect to x, as the gradient of a curve is given by the derivative.

Differentiating the equation y px qy = 0 with respect to x:

d/dx (y px qy) = d/dx (0)

Using the product rule, we differentiate each term separately:

(pxy + py dx/dx + qy dx/dx) = 0

Simplifying the equation:

pxy + py + qy dx/dx = 0

Since we are given that the point (1, 1) lies on the curve, we can substitute x = 1 and y = 1 into the equation:

p(1)(1) + p(1) + q(1) = 0

p + p + q = 0

2p + q = 0 ---(1)

Now, we need to find the value of dx/dy to calculate the gradient at the point (1, 1). We can differentiate the equation with respect to y:

d/dy (y px qy) = d/dy (0)

Using the product rule again, we differentiate each term separately:

(px dy/dy + py y' + qy dy/dy) = 0

Simplifying the equation:

px + py y' + qy = 0

Since we know that y = 1 at the point (1, 1), we can substitute y = 1 into the equation:

px + py y' + qy = 0

px + py y' + q(1) = 0

px + py y' + q = 0

Substituting x = 1 and y = 1:

p + py' + q = 0

Since we are given that the gradient at (1, 1) is 1/2, we can substitute y' = 1/2:

p + p(1/2) + q = 0

p/2 + p + q = 0

3p/2 + q = 0 ---(2)

Solving the equations:
We have two equations:
2p + q = 0 ---(1)
3p/2 + q = 0 ---(2)

Multiplying equation (2) by 2 to eliminate fractions:

3p + 2q = 0 ---(3)

Now we have two equations with two unknowns (p and q). We can solve these equations simultaneously to find the values of p and q.

Using equation (1) and equation (3):

2p + q = 0 ---(1)
3p + 2q = 0 ---(3)

Multiplying equation (1) by 2:

4p + 2q = 0 ---(4)

Subtracting equation (4) from equation (3):

3p + 2q - (4p + 2q) = 0

3p + 2q - 4p - 2q = 0

-p =
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The gradient of the curve y px qy=0 at (1,1) is1/2.the values of P and q are?
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