What is 4x\mid ^2+y^2<16\mid ?

The solution to 4x\mid ^2+y^2<16\mid is Parabola with (h,k)=(4/(\mid),0),p=-\mid ^2

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4x\mid ^2+y^2<16\mid

4 x ∣^{2} + y^{2} < 16 ∣

(h, k) = (\frac{4}{∣}, 0), p = - ∣^{2}

Solution steps

4 x ∣^{2} + y^{2} < 16 ∣

Rewrite

4 x ∣^{2} + y^{2} < 16 ∣

in the standard form:

4 (- ∣^{2}) (x - \frac{4}{∣}) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (\frac{4}{∣}, 0), p = - ∣^{2}

9 x 6 + 16 y^{2} = 144

y^{2} - 12 x - 2 y + 25 = 0

y = - 3 x^{2} - 6 x - 12

2 y^{2} + 8 y - x + 5 = 0

5^{2} + 5 y^{2} + 10 x + 2 y + 3 = 0

What is 4x\mid ^2+y^2<16\mid ?
The solution to 4x\mid ^2+y^2<16\mid is Parabola with (h,k)=(4/(\mid),0),p=-\mid ^2