The sum of the reciprocals of the focal dista | vedclass

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

Question

(A) Let the coordinates of points $P$ and $Q$ on the parabola $y^{2} = 4ax$ be $(at_{1}^{2}, 2at_{1})$ and $(at_{2}^{2}, 2at_{2})$ respectively.Since

Vedclass · Accepted Answer

$\frac{1}{a}$

Vedclass · Answer

(A) Let the coordinates of points $P$ and $Q$ on the parabola $y^{2} = 4ax$ be $(at_{1}^{2}, 2at_{1})$ and $(at_{2}^{2}, 2at_{2})$ respectively.Since $PQ$ is a focal chord,$t_{1}t_{2} = -1$.The focal distances of points $P$ and $Q$ are $r_{1} = a(1 + t_{1}^{2})$ and $r_{2} = a(1 + t_{2}^{2})$.The sum of the reciprocals is $\frac{1}{r_{1}} + \frac{1}{r_{2}} = \frac{1}{a(1 + t_{1}^{2})} + \frac{1}{a(1 + t_{2}^{2})}$.Substituting $t_{2} = -\frac{1}{t_{1}}$,we get $\frac{1}{a(1 + t_{1}^{2})} + \frac{1}{a(1 + \frac{1}{t_{1}^{2}})} = \frac{1}{a(1 + t_{1}^{2})} + \frac{t_{1}^{2}}{a(t_{1}^{2} + 1)}$.$= \frac{1 + t_{1}^{2}}{a(1 + t_{1}^{2})} = \frac{1}{a}$.

The sum of the reciprocals of the focal distances of a focal chord $PQ$ of the parabola $y^{2} = 4ax$ is:

Similar Questions

The latus rectum of a parabola whose focal chord $PSQ$ is such that $SP = 3$ and $SQ = 2$ is given by

The shortest distance from $(0,3)$ to the parabola $y^2=4x$ is

The equation of the tangent to the parabola $y^2=16x$,which is perpendicular to the line $3x-4y+5=0$,is given by

The equation of the tangent to the parabola $y^2=12x$,which makes an angle $30^{\circ}$ with the positive direction of the $X$-axis is given by $x-\sqrt{3}y+9=0$. The point of contact is:

If $ax^2+2hxy+by^2-82x+98y+144=0$ is the equation of a parabola with focus $(2,-3)$ and directrix $3x-2y+5=0$,then $ax^2+2hxy+by^2=0$ represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module