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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The equation of a circle is \(\operatorname{Re}\left(z^{2}\right)+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0\), where \(z=x+\) iy. A line which passes through the center of the given circle and the vertex of the parabola, \(x^{2}-6 x-y+13=0,\) has \(y\)-intercept equal to \(.....\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
Equation of circle is \(\left(x^{2}-y^{2}\right)+2 y^{2}+2 x=0\) \(X^{2}+y^{2}+2 x=0\) Centre \(:(-1,0)\) Parabola \(: x^{2}-6 x-y+13=0\) \((x-3)^{2}=y-4\) Vertex: \((3,4)\) \(\equiv \,y-0=\frac{4-0}{3+1}(x+1)\) \(y=x+1\) \(y\) - intercept \(=1\)
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