A parabola has its focus on the positive X-ax | vedclass

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(D) Let the focus be $(a, 0)$ and the directrix be $x = 0$. The parabola is defined by the locus of points $(x, y)$ such that the distance to $(a, 0)$

Vedclass · Accepted Answer

$\sqrt{32}$

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(D) Let the focus be $(a, 0)$ and the directrix be $x = 0$. The parabola is defined by the locus of points $(x, y)$ such that the distance to $(a, 0)$ equals the distance to $x = 0$: $(x-a)^2 + y^2 = x^2$. Simplifying, $y^2 = 2ax - a^2 = 2a(x - a/2)$.Since $P(\alpha, 4)$ lies on the parabola, $16 = 2a(\alpha - a/2) = 2a\alpha - a^2$.The equation of the tangent at $P(\alpha, 4)$ to the parabola $y^2 = 2ax - a^2$ is $4y = a(x + \alpha) - a^2$.Since the tangent passes through $(0, 0)$, we have $0 = a\alpha - a^2$, which implies $a = \alpha$ (since $a 
eq 0$).Substituting $a = \alpha$ into the parabola equation: $16 = 2\alpha^2 - \alpha^2 = \alpha^2$.Thus, $\alpha^2 = 16$, so $\alpha = 4$ (since the focus is on the positive $X$-axis, $\alpha > 0$).The point $P$ is $(4, 4)$.The distance of $P$ from the origin is $\sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}$.

$A$ parabola has its focus on the positive $X$-axis and the $Y$-axis as its directrix. If $P(\alpha, 4)$ is a point on this parabola such that the tangent to the parabola at point $P$ passes through the origin, then the distance of $P$ from the origin is

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