The maximum number of points on the parabola | vedclass

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(C) Let the point $P$ be $(h, k)$. $A$ point $(x, y)$ on the parabola $y^2 = 16x$ is equidistant from $P$ if it lies on a circle centered at $P$ with

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the point $P$ be $(h, k)$. $A$ point $(x, y)$ on the parabola $y^2 = 16x$ is equidistant from $P$ if it lies on a circle centered at $P$ with radius $r$. Thus,the distance condition is $(x - h)^2 + (y - k)^2 = r^2$. Substituting $x = \frac{y^2}{16}$ into the circle equation,we get $(\frac{y^2}{16} - h)^2 + (y - k)^2 = r^2$. This simplifies to a quartic equation in $y$: $\frac{y^4}{256} - \frac{hy^2}{8} + h^2 + y^2 - 2ky + k^2 = r^2$,which is $\frac{y^4}{256} + (1 - \frac{h}{8})y^2 - 2ky + (h^2 + k^2 - r^2) = 0$. $A$ quartic equation can have at most $4$ real roots. Therefore,a circle can intersect a parabola at a maximum of $4$ points.

The maximum number of points on the parabola $y^2 = 16x$ that are equidistant from a variable point $P$ (which lies inside the parabola) is -

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