For the parabola y^2 = 8(x - 3),let P be a po | vedclass

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

Question

(D) The equation of the parabola is $y^2 = 8(x - 3)$.Comparing this with $(y - k)^2 = 4a(x - h)$,we get $4a = 8$,so $a = 2$.The vertex is $(h, k) = (3

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The equation of the parabola is $y^2 = 8(x - 3)$.Comparing this with $(y - k)^2 = 4a(x - h)$,we get $4a = 8$,so $a = 2$.The vertex is $(h, k) = (3, 0)$ and the focus $S$ is $(h + a, k) = (3 + 2, 0) = (5, 0)$.The directrix is $x = h - a = 3 - 2 = 1$.Let $P$ be a point on the parabola. By definition,the distance $SP$ is equal to the perpendicular distance from $P$ to the directrix,which is $PM$.Since $	riangle SPM$ is an equilateral triangle,$SP = PM = SM$.Let $M$ be the point $(1, y_P)$ on the directrix. The distance $SM$ is the distance between $(5, 0)$ and $(1, y_P)$,so $SM = \sqrt{(5 - 1)^2 + (0 - y_P)^2} = \sqrt{16 + y_P^2}$.Also,in the right-angled triangle formed by the projection of $P$ onto the directrix,the angle $\angle SMP = 60^\circ$ (as it is equilateral).Let $Z$ be the point $(1, 0)$ on the directrix. In $	riangle SZM$,$\angle SZM = 90^\circ$ and $SZ = 5 - 1 = 4$.Since $\angle SMZ = 30^\circ$,we have $\cos(30^\circ) = \frac{SZ}{SM}$.$\frac{\sqrt{3}}{2} = \frac{4}{SM} \Rightarrow SM = \frac{8}{\sqrt{3}}$.Wait,looking at the geometry,$PM$ is horizontal. The distance $PM$ is the horizontal distance from $x=1$ to $P(x, y)$,so $PM = x - 1$.Since $SP = PM$,$SP = x - 1$. Also $SP = x + a - 2h = x + 2 - 6 = x - 4$ is incorrect; $SP = x + a = x + 2$ relative to vertex. Actually $SP = x - 3 + 2 = x - 1$.In $	riangle SPM$,$PM = SP = SM = L$. The height of the triangle from $S$ to $PM$ is $L \sin(60^\circ) = \frac{L\sqrt{3}}{2}$.The vertical distance from $S(5, 0)$ to the line $PM$ is $y_P$. So $y_P = \frac{L\sqrt{3}}{2}$.Also $PM = x_P - 1 = L$. Since $y_P^2 = 8(x_P - 3)$,we have $(\frac{L\sqrt{3}}{2})^2 = 8(L + 1 - 3) = 8(L - 2)$.$\frac{3L^2}{4} = 8L - 16 \Rightarrow 3L^2 - 32L + 64 = 0$.$(3L - 8)(L - 8) = 0$. Since $L$ must be $8$ for the geometry to hold,$L = 8$.

For the parabola $y^2 = 8(x - 3)$,let $P$ be a point on it. Let $M$ be the foot of the perpendicular from $P$ to the directrix,and $S$ be the focus of the parabola. If $\triangle SPM$ is an equilateral triangle,find the length of each side of the triangle.

Similar Questions

The length of the latus rectum of the parabola whose focus is at $(1, -2)$ and directrix is the line $x + y + 3 = 0$ is

Let $P$ represent the point $(3, 6)$ on the parabola $y^2 = 12x$. For the parabola $y^2 = 12x$,if $l_1$ is the length of the normal chord drawn at $P$ and $l_2$ is the length of the focal chord drawn through $P$,then $\frac{l_1}{l_2} = $

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2}=4x$ be the curve $S$. Let $P$ be any point on $S$. Then the locus of the point, which internally divides $OP$ in the ratio $3:1$, is:

The point on the parabola $2y = x^2$ which is nearest to the point $(0, 3)$ is

Let $P, Q,$ and $R$ be three co-normal points on the parabola $y^2 = 4ax$. Then the correct statement$(s)$ is/are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module