The given circle x^2 + y^2 + 2px = 0,p in R t | vedclass

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(B) The equation of the parabola is $y^2 = 4x$,which opens to the right with its vertex at $(0, 0)$.The equation of the circle is $x^2 + y^2 + 2px = 0

Vedclass · Accepted Answer

$p > 0$

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(B) The equation of the parabola is $y^2 = 4x$,which opens to the right with its vertex at $(0, 0)$.The equation of the circle is $x^2 + y^2 + 2px = 0$,which can be rewritten as $(x + p)^2 + y^2 = p^2$.This represents a circle with center $C = (-p, 0)$ and radius $r = |p|$.For the circle to touch the parabola externally,the center of the circle must lie on the negative $x$-axis (since the parabola is in the region $x \ge 0$).If $p > 0$,the center is $(-p, 0)$,which is on the negative $x$-axis. The circle lies to the left of the $y$-axis and touches the parabola at the origin $(0, 0)$ externally.If $p Thus,the condition for external contact is $p > 0$.

The given circle $x^2 + y^2 + 2px = 0$,$p \in R$ touches the parabola $y^2 = 4x$ externally,then

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