Home
Class 12
MATHS
ax^2+2hxy+by^2+2gx+2fy+c=0...

`ax^2+2hxy+by^2+2gx+2fy+c=0`

Text Solution

Verified by Experts

`ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`
Differentiating w.r.t. x, we get,
`2ax+2h(x(dy)/(dx)+y)+2by(dy)/(dx)+2g+2f.(dy)/(dx)+0=0`
Cancelling 2 throughout, we get,
`ax+hx(dy)/(dx)+hy+by(dy)/(dx)+g+f.(dy)/(dx)=0`
`(hx+by+f)(dy)/(dx)=-ax-hy-g`
`therefore (dy)/(dx)=-((ax+hy+g))/(hx+by+f)`
Promotional Banner

Similar Questions

Explore conceptually related problems

If the system of equations ax+hy+g=0 ….(i) hx+by+f=0…(ii) and ax^2+2hxy+by^2 + 2gx+ 2fy+c+t=0 ….(iii) has a unique solution and (abc+2fgh-af^2-bg^2-ch^2)/(h^2-ab)=8 , find the value of 't'.

the equation ax^(2)+ 2hxy + by^(2) + 2gx + 2 fy + c=0 represents an ellipse , if

For the equation ax^(2) +by^(2) + 2hxy + 2gx + 2fy + c =0 where a ne 0 , to represent a circle, the condition will be

If ax^(2)+2hxy+by^(2)+2gx+2fy+c-=(lx+my+n)(l'x+m'y+n') , then prove that : |{:(a,h,g),(h,b,f),(g,f,c):}|=0

Equation of a circle that cuts the circle x^2 + y^2 + 2gx + 2fy + c = 0 , lines x=g and y=f orthogonally is : (A) x^2 + y^2 - 2gx - 2fy - c = 0 (B) x^2 + y^2 - 2gx - 2fy - 2g^2 - 2f^2 - c =0 (C) x^2 + y^2 + 2gx + 2fy + g^ + f^2 - c = 0 (D) none of these

If ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 represents an ellipse, then h^2 lt ab and abc +2fgh-at^2-bg^2-ch^2 != 0. If for every point (x_1,y_1) satisfying above equation onthen (2h-x_1,2k-y_1) also satisfy it,then (h,k) is centre of it. The length of semi major axis andmaximum and minimum value of the distance of a point lying on the curve from its centre..For the ellipse 2x^2 - 2xy + 4y^2 - (3 + sqrt2) = 0, the bindination of major axis of it with x-axis is