If the circle x^2+y^2+2 g x+2 f y+c=0 cuts the curve x=p t, y=(p)/(t)(t being a parameter) at A,

If the circle x^2+y^2+2 g x+2 f y+c=0 cuts the curve x=p t,...

Key Concepts

Circle Equation

The circle given in the problem is expressed in its general quadratic form, which represents a circle in the Cartesian plane. Understanding the circle's equation is vital because it provides the basis for establishing geometric relationships and intersections. In general, manipulating the circle's equation allows one to explore its properties such as center, radius, and intersection points with other curves.

Parameterization of Curves

Parameterization is a method of expressing the coordinates of the points on a curve as functions of a parameter. In this question, the curve is parameterized by t with x = p*t and y = p/t. This approach simplifies the process of finding intersection points with the circle by reducing the problem to solving equations in a single variable.

Intersection of Curves via Substitution

When a circle and a parameterized curve intersect, the coordinates of the intersection points must satisfy both equations simultaneously. Substituting the parameterized expressions for x and y into the circle's equation leads to a polynomial in the parameter t. This method is a standard analytical technique to reduce a system of equations to a single variable, making the analysis of the intersection points manageable.

Vieta's Formulas

Vieta's formulas relate the coefficients of a polynomial to the sums and products of its roots. In the context of this problem, once the substitution yields a polynomial in t, Vieta's formulas are employed to express complex relationships—like the product of the roots (parameters corresponding to the intersection points) and the sum of their reciprocals—in terms of the circle's and curve's given constants. This powerful algebraic tool simplifies the derivation of the required relationships between the intersection parameters.

Transcript

00:01 In this problem of conic section we have given that if the circle whose equation is x square plus y square plus 2g x plus 2 f y plus c is equal to 0 cuts the curve x is equal to p t and y is equal to p divided with t and t being a parameter at a b c and d with parameters say a b c and d with some parameters which is given why this is t1 t 2 t 3 and t 4 respectively then we have to prove that t 1 multiplied with t 2 multiplied with t 2 multiplied with t 4 is equal to 1 so first we have the equation of circle and here this is the hyperbola so hyperbola is x y is equals to p square so this is the hyperbola and now this is point a this is t 1 so we can can say here the points would be say p t 1 and p divided with t 1 similarly this would be p t 2 and p divided with t 2 and similarly here this would be p t 3 and divided with pt 3 and here this would be be t 4 and p divided with t 4 so these are the 4 so let's p t and pt divided with t is a point so let's pt and p point on is a point on s so here we have given s is equal to this equation or we can say s is the equation of this circle and now from here when we put this value x is equals to p t y is equal to p divided with t so p t whole square is p d divided with two whole square is p squared d divide with t square and twice of g p t plus twice of f p divided with t and plus c is equal to zero now when we solve it so we are getting this is p square and here we are multiplying t square to all terms so this would be p square and t to the power four plus p square plus two g p cube so this would be here p not so this would be p and t cube actually and here this would be twice of f p multiplied with t and c t square is equals to 0 so we can say this is equation number 1 and from here we can say so from here we can say that t1 multiplied with t2 multiplied with t 3 and t 4 is basically equals to 1 and now for part 2 we have to prove that 1 divided with this is 1 divided with t 1 plus 1 divided with t 2 plus 1 divided with t 3 plus 1 divided with t 4 is equal to 2 f divided with p so now we can reciproc take the reciprocal of equation number 1 so we are take reciprocal of equation number 1 so this is your equation number 1 and this would be p square t to the power 4 plus twice of f p t q plus 2 g p t plus p t plus p is equal to 0 so from here we can say this is 1 divided with t 1 plus 1 divided with t2 plus 1 divided with t3 plus 1 divided with t4 is equals to minus 2 f divided with pt...

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