ParabolaMedium
Question
Let P be a moving point on the circle
$$x^{2} + y^{2} - 6x - 8y + 21 = 0$$
Then, the maximum distance of P from the vertex of the parabola $x^{2} + 6x + y + 13 = 0$ is equal to :
Options
A.8
B.10
C.12
D.9
Solution
Sol. [Key Concept: Maximum distance $=$ (distance from the circle's centre to the parabola's vertex) $+$ radius.]
Centre : $C(3,4)$ & $r = 2$
Parabola $(x + 3)^{2} = - (y + 4)$
Vertex is $A( - 3, - 4)$
$${AP}_{\text{max}\text{~}} = AC + r = \sqrt{36 + 64} + 2 = 12 $$
Create a free account to view solution
View Solution FreeMore Parabola Questions
$Px + 2y = 1$ is normal to parabola $y^{2} = 4ax$ for :...The equation of the circle drawn with the focus of the parabola (x - 1)2 - 8y = 0 as its centre and touching the parabol...Through the vertex O of the parabola, y2 = 4ax two chords OP and OQ are drawn and the circles on OP and OQ as diameter i...From the point (4, 6) a pair of tangent lines are drawn to the parabola, y2 = 8x. The area of the triangle formed by the...Let the directrix of the parabola $P:y^{2} = 8x$, cut x -axis at the point A . Let $B(\alpha,\beta),\alpha > 1$, be a...