Let the point P be the vertex of the parabola | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the parabola is $y = (x-3)^2 + 3$, so the vertex $P$ is $(3, 3)$.The equation of the circle is $(x-1)^2 + (y-2)^2 = 2$, with cente

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The equation of the parabola is $y = (x-3)^2 + 3$, so the vertex $P$ is $(3, 3)$.The equation of the circle is $(x-1)^2 + (y-2)^2 = 2$, with center $O(1, 2)$ and radius $r = \sqrt{2}$.The distance $PO = \sqrt{(3-1)^2 + (3-2)^2} = \sqrt{4+1} = \sqrt{5}$.Let the line through $P$ make an angle $	heta$ with the line $PO$. Let $d$ be the perpendicular distance from $O$ to the line $RS$. Then $d = PO \sin 	heta = \sqrt{5} \sin 	heta$.The lengths $PR$ and $PS$ are the roots of the quadratic equation obtained by substituting the line equation into the circle equation. If $d_1, d_2$ are the distances $PR$ and $PS$, then $d_1 + d_2 = 2 \sqrt{r^2 - d^2} \cos \phi$ is not the standard approach. Instead, using the property of the chord, $PR + PS = 2 \sqrt{PO^2 - d^2} \cos \alpha$ where $\alpha$ is the angle between $PO$ and the line. More simply, $PR$ and $PS$ are $d \pm \sqrt{r^2 - d^2}$. Thus $PR+PS = 2 \sqrt{PO^2 - d^2} = 2 \sqrt{5 - d^2}$.To maximize $(PR+PS)^2 = 4(5 - d^2)$, we minimize $d^2$. The minimum value of $d$ is $0$ (when the line passes through the center $O$).Thus, the maximum value is $4(5 - 0) = 20$.

Let the point $P$ be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point $P$ intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points $R$ and $S$, then the maximum value of $(PR + PS)^2$ is:

Similar Questions

The equation of a circle touching the parabola $y = x^2$ at the point $(1, 1)$ and passing through the point $(2, 2)$ is:

The locus of the centers of the circles such that the point $(2, 3)$ is the midpoint of the chord $5x + 2y = 16$ is:

If the line $x \cos \theta + y \sin \theta = 2$ is the equation of a transverse common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6 \sqrt{3} x - 6y + 20 = 0$,then the value of $\theta$ is:

$A$ circle touches the line $2x + y - 10 = 0$ at $(3, 4)$ and passes through the point $(1, -2)$. Then a point that lies on the circle is

If $\theta$ is the acute angle of intersection at a real point of intersection of the circle $x^2 + y^2 = 5$ and the parabola $y^2 = 4x$,then $\tan \theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module