If two curves whose equations are $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ and $a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x + 2 f^{'} y + c^{'} = 0$ intersect in four concyclic points, prove that $\frac{a - b}{h} = \frac{a^{'} - b^{'}}{h^{'}}$ .

Open in App

Solution

The equation of the conic through the four points of intersection of $S_{1} = 0$ and $S_{2} = 0$ is given by $S_{1} - λ S_{2} = 0$ .
If it be a circle, then
coeff. of $x^{2} = c o e f f . o f y^{2} ∴ a - λ a^{'} = b - λ b^{'}$
or $a - b = λ (a^{'} - b^{'}) . . . . . (1)$
Coeff. of $x y = 0 ∴ 2 h - λ . 2 h^{'} = 0$
$h = λ h^{'} . . . . . . (2)$
Eliminating $λ$ , we get $\frac{a - b}{h} = \frac{a^{'} - b^{'}}{h^{'}}$ .

Suggest Corrections

Similar questions

Q. If two curves whose equations are

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

and

a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x + 2 f^{'} y + c^{'} = 0

intersect in four concyclic points, then -

Q. lf two conics whose equations refer to rect- angular axes are

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

and

a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x + 2 f y + c^{'} = 0

intersect in four concyclic points, then

Q. Show that the straight lines joining the origin to the other two points of intersection of the curves whose equations are

a x^{2} + 2 h x y + b y^{2} + 2 g x = 0

and

a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x = 0

will be at right angles if

g (a^{'} + b^{'}) - g^{'} (a + b) = 0 .

Q. The straight line joining the origin to the other two points of intersection of the curve whose equations are

a x^{2} + 2 h x y + 2 g x + b y^{2} = 0 a n d a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x = 0

will be at right angle if

The line joining the origin to the points of intersection of the curves $a x^{2} + 2 h x y + b y^{2} + 2 g x = 0$ and $a^{^{'}} x^{2} + 2 h^{^{'}} x y + b^{^{'}} y^{2} + 2 g^{^{'}} x = 0$ will be mutually perpendicular, if

Join BYJU'S Learning Program

If two curves whose equations are ax2+2hxy+by2+2gx+2fy+c=0 and a′x2+2h′xy+b′y2+2g′x+2f′y+c′=0 intersect in four concyclic points, prove that a−bh=a′−b′h′.

If two curves whose equations are $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ and $a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x + 2 f^{'} y + c^{'} = 0$ intersect in four concyclic points, prove that $\frac{a - b}{h} = \frac{a^{'} - b^{'}}{h^{'}}$ .