Question

The locus of the point of intersection of thr straight lines tx-2y-3t =0 and x-2ty+y3=0 is

Answer 1

Answer:

x² - 4y² = 9.

Step-by-step explanation:

Hi,

Given straight lines are

tx - 2y =3t-----------------(1)

x  - 2ty = -3---------------(2)

Finding the point of intersection (h, k) if the above 2 lines we get

(t² - 1)x = 3(t² + 1)

=> h = 3(t² + 1)/(t² - 1)

=> h/3 = (t² + 1)/(t² - 1)

2(t² - 1)y = 6t

=> k = 3t/(t² - 1)

=>2k/3 = 2t/(t² - 1)

Consider (h/3)² - (2k/3)²

= 1/(t² - 1)²[(t² + 1)² - 4t²]

= 1/(t² - 1)²*(t² - 1)²

=1

Hence , the locus of point of intersection of the given 2 lines is

x² - 4y² = 9.

Hope, it helped !

Answer 2

Answer:

Step-by-step explanation:

Concept:

Locus:

The path of the moving point according to some geometrical conditions is called locus.

Given lines are

tx-2y-3t =0 and x-2ty+y3=0

Let P(h,k) be the point of intersection of given two lines.

Then

th-2k-3t=0 and h-2tk+3=0

The locus of P is obtained by eliminating t from these two equations.

t(h-3)=2k and 2tk=h+3

$t=\frac{2k}{h-3}\:and\:t=\frac{h+3}{2k}$

This implies

$\frac{2k}{h-3}=\frac{h+3}{2k}\\\\4k^2=(h+3)(h-3)\\\\4k^2=h^2-9\\\\h^2-4k^2=9\\\\\frac{h^2}{9}-\frac{k^2}{9/4}=1$

Therefore the locus of P is

$\frac{x^2}{9}-\frac{y^2}{9/4}=1$

This equation represents a hyperbola.