If the equation $k y^{2} + y = x^{2} - 16 x + 64$ represents a parabola then the value of $| 4 Δ | =$

Open in App

Solution

The given equation is $k y^{2} + y = x^{2} - 16 x + 64$
compairing it with standard conic equation,
we get $a = 1, b = - k, h = 0, g = - 8, f = - 1 / 2, c = 64$
For equation to be parabola
$h^{2} = a b \Rightarrow k = 0$
Now
$Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0 + 0 - 1 / 4 = - \frac{1}{4} \Rightarrow | 4 Δ | = 1$

Suggest Corrections

Similar questions

k y^{2} + y = x^{2} - 16 x + 64

| 4 Δ | =

k y^{2} + y = x^{2} - 16 x + 64

Δ

| 4 Δ |

k y^{2} + y = x^{2} - 16 x + 64

Δ

| 4 Δ |

α, β

x^{2} - 2 x + 4 = 0

α^{3}, β^{3}

8

x -

y -

Join BYJU'S Learning Program