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Find the equation of the conic which passes through the points $(-1,-1)$ and also through the intersection of the conic $x^2+2xy+5y^2+x+16y+8=0$ with the straight lines $2x-y-3=0$ and $3x+y-3=0$. i failed to solve this question.the solution provided by the book is:$$34x^2+55xy+139y^2+13x+348y+233=0$$Can anyone help me to explain how to get the solution.

Thanks in advance.

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    $\begingroup$ Plugging each of the points into the general conic equation gives you a system of linear equations in the unknown coefficients. $\endgroup$ – amd Aug 30 '18 at 20:49
  • $\begingroup$ but how to get the intersection point of the conic $x^2+2xy+5y^2+x+16y+8=0$ with the straight lines $2x-y-3=0$ and $3x+y-3=0$ @amd $\endgroup$ – emonHR Aug 30 '18 at 20:52
  • $\begingroup$ You wrote in your question that you know how to do that. $\endgroup$ – amd Aug 30 '18 at 20:53
  • $\begingroup$ Also, check the book solution: $348y$ is incorrect. $\endgroup$ – amd Aug 30 '18 at 20:56
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The union of the two straight lines is a degenerate conic, with equation $$(2x-y-3)(3x+y-3)=0.$$ The conic which passes through the points of intersection of these two conics belongs to the pencil of conics with equation $$\lambda(x^2+2xy+5y^2+x+16y+8)+\mu(2x-y-3)(3x+y-3)=0$$ for some $\lambda, \mu$. You just have to write this conic passes through $(-1,-1)$ to determine the ratio $(\lambda:\mu)$ (this pencil of conics is a projective line).

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Each of straight lines cuts the given conic in 2 points, the fifth point $(-1,-1)$ is given. Plug the coordinates of these points into the equation $$ax^2+bxy+cy^2+dx+ey+f=0$$ of the resulting conic. Then solve the system of 5 linear equations with unknowns $a,b,c,d,e,f$ (one of the will be a parameter).

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