The locus of the midpoint of the focal radii | vedclass

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

Question

(D) Let the variable point on the parabola $y^2 = 4ax$ be $P(at^2, 2at)$.The focus of the parabola is $S(a, 0)$.The focal radius is the segment $SP$.

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) Let the variable point on the parabola $y^2 = 4ax$ be $P(at^2, 2at)$.The focus of the parabola is $S(a, 0)$.The focal radius is the segment $SP$. The midpoint $M(x, y)$ of $SP$ is given by:$x = \frac{at^2 + a}{2}, y = \frac{2at + 0}{2} = at$.From $y = at$,we get $t = \frac{y}{a}$.Substituting $t$ into the equation for $x$:$x = \frac{a(\frac{y}{a})^2 + a}{2} = \frac{\frac{y^2}{a} + a}{2} = \frac{y^2 + a^2}{2a}$.$2ax = y^2 + a^2 \implies y^2 = 2ax - a^2 = 2a(x - \frac{a}{2})$.This is a parabola with vertex at $(\frac{a}{2}, 0)$.The latus rectum of this new parabola is $4A = 2a$,which is half the latus rectum of the original parabola $(4a)$.The directrix is $x - \frac{a}{2} = -\frac{a}{2} \implies x = 0$,which is the $y$-axis.Thus,all the given statements are correct.

The locus of the midpoint of the focal radii of a variable point moving on the parabola $y^2 = 4ax$ is a parabola whose:

Similar Questions

Let $A(4, -4)$ and $B(9, 6)$ be points on the parabola $y^2 = 4x$. Let $C$ be a point chosen on the arc $AOB$ of the parabola,where $O$ is the origin,such that the area of $\Delta ACB$ is maximum. Then,the area (in sq. units) of $\Delta ACB$ is

If $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a focal chord of the parabola $y^2 = 4ax$,then what is the square of the $G.M.$ of $x_1$ and $x_2$?

Find the minimum distance from the point $(0, c)$ to the parabola $y = x^2$,where $0 \leq c \leq 5$.

The equation of the tangent at $(-4, -4)$ on the curve $x^2 = -4y$ is

The directrix of the parabola $2 y^2+25 x=0$ is $........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module