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Q.
The slope of the straight line joining the centre of the circle $ x^2 + y^2-8x + 2y = 0$ and the vertex of the parabola $ y = x^2 - 4x + 10 $ is
Conic Sections
Solution:
Centre of the circle $x^{2}+y^{2}-8x+2y =0$ is $\left(4, -1\right)$
Given parabola is $y= \left(x-2\right)^{2} +6 $
$ \Rightarrow \left(x-2\right)^{2}=4\cdot\frac{1}{4}\left(y-6\right)$
Vertex is $\left(2, 6\right)$
Required slope $= \left(\frac{6}{2} \,\, \frac{1}{4}\right)$
$ =- \frac{7}{2}$