Question

please help me

find solutions for 2 bit​

Answer 1

Step-by-step explanation:

Question :-

The locus given by :-

$16 {x}^{2} - 24xy + 9 {y}^{2} - 62x + 34y + 46 = 0$

Solution :-

Comparing :-

$16 {x}^{2} - 24xy + 9 {y}^{2} - 62x + 34y + 46 = 0$

with , General form of Locus equation (S):-

$a {x}^{2} + b {y}^{2} + 2hxy + 2gx + 2fy + c = 0$

$\sf\implies$

1) a = 16

2) b = 9

3)2h = -24 , h = -24/2 = -12

4) 2g = -62 , g = -62/2 = - 31

5) 2f = 34 , f = 34/2 = 17

6) c = 46

We know that :-

$\triangle \: = abc + 2fgh - a {f}^{2} - b {g}^{2} - c {h}^{2}$

= 16(9)(46) + 2(17)(-31)(-12) - 16(17)^2 - 9(-31)^2 - 46(-12)^2

= 144(46) + 34(372) - 16(289) - 9(961) - 46(144)

= 6624 + 12648 - 4624 - 8649 - 6624

= 12648−13273

= - 625

$\triangle \: = - 625$

$\implies \: \triangle \: ≠0$

h^2 = (-12)^2 = 144

ab = 16(9) = 144

$\implies \: {h}^{2} = ab$

If $\sf\triangle ≠ 0 \: and\: h^{2} = ab$ Then that line is "Parabola"