Question

the locus of the point (4t^2 -1 , 8t-2 ) represents a parabola then the equation of latus rectum is???
please give solution

Answer 1

Answer:

Equation of latusrectum is x=3

Step-by-step explanation:

Given:

$x=4t^2-1$

$y=8t-2$

This can be written as

$x+1=4t^2$

$y+2=8t$

squaring on both sides

$(y+2)^2=64t^2$

$(y+2)^2=16(4t^2)$

$(y+2)^2=16(x+1)$

$Y^2=16X$

where X= x+1 and Y=y+2

This equation of the form

$y^2=4ax$ (standard equation of open rightward parabola )

Therefore the equation

$(y+2)^2=16(x+1)$ represents

open rightward parabola.

Here

4a = 16

a=4

Equation of latus rectum is

X=a

x+1=4

x=3