If one end of a focal chord of the parabola y | vedclass

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

Question

(C) The given parabola is $y^2=8x$. Comparing this with the standard form $y^2=4ax$,we get $4a=8$,so $a=2$. $A$ point on the parabola is given by $(at

Vedclass · Accepted Answer

$\frac{25}{2}$

Vedclass · Answer

(C) The given parabola is $y^2=8x$. Comparing this with the standard form $y^2=4ax$,we get $4a=8$,so $a=2$. $A$ point on the parabola is given by $(at^2, 2at) = (2t^2, 4t)$. Given one end of the focal chord is $\left(\frac{1}{2}, 2ight)$,we have $4t=2$,which implies $t=\frac{1}{2}$. The length of a focal chord with parameter $t$ is given by the formula $L = a\left(t + \frac{1}{t}ight)^2$. Substituting $a=2$ and $t=\frac{1}{2}$: $L = 2\left(\frac{1}{2} + \frac{1}{1/2}ight)^2 = 2\left(\frac{1}{2} + 2ight)^2$. $L = 2\left(\frac{5}{2}ight)^2 = 2 	imes \frac{25}{4} = \frac{25}{2}$ units.

If one end of a focal chord of the parabola $y^2=8x$ is $\left(\frac{1}{2}, 2\right)$,then the length of the focal chord is $........$ units.

Similar Questions

The distance of the point $(6, -2 \sqrt{2})$ from the common tangent $y = mx + c$ $(m > 0)$ of the curves $x = 2y^2$ and $x = 1 + y^2$ is

At what points does the line $2x + y - 1 = 0$ intersect the parabola $y^2 = 4x$?

Tangent and normal are drawn at $P(16, 16)$ on the parabola ${y^2} = 16x$,which intersect the axis of the parabola at $A$ and $B$,respectively. If $C$ is the centre of the circle through the points $P, A$ and $B$ and $\angle CPB = \theta$,then a value of $\tan \theta$ is:

If $lx + my + n = 0$ is tangent to the parabola $x^2 = y$,then the condition of tangency is

The equation of the lines joining the vertex of the parabola $y^2 = 6x$ to the points on it whose abscissa is $24$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module