If-the-system-y-mx-2-2-4x-2-y-2-4-have-only-one-solution-them-m-A-1-3-B-1-2-C-1-D-2-E-3-

Question Number 207866 by efronzo1 last updated on 29/May/24

If the system {\begin{cases} y = - {mx}^{2} - 2 \\ {4 x}^{2} + y^{2} = 4 \end{cases}

have only one solution

them m =

(A) \frac{1}{3} (B) \frac{1}{\sqrt{2}} (C) 1 (D) \sqrt{2} (E) \sqrt{3}

Answered by Rasheed.Sindhi last updated on 29/May/24

{\begin{cases} y = - {mx}^{2} - 2 \dots (i) \\ {4 x}^{2} + y^{2} = 4 \dots (i i) \end{cases}

x^{2} = \frac{- y - 2}{m} from (i)

putting in (ii) :

4 (\frac{- y - 2}{m}) + y^{2} - 4 = 0

{my}^{2} - 4 m - 4 y - 8 = 0

{my}^{2} - 4 y - 4 m - 8 = 0

y = \frac{4 \pm \sqrt{16 - 4 m (- 4 m - 8)}}{2 m}

For one solution

16 + {16 m}^{2} + 32 m = 0

m^{2} + 2 m + 1 = 0

{(m + 1)}^{2} = 0

m = - 1

Commented by A5T last updated on 29/May/24

m = 0 a l s o w o r k s .

m = 0 \Rightarrow y = - 2 \Rightarrow x = 0 \Rightarrow u n i q u e (x, y) = (0, - 2)

Answered by A5T last updated on 29/May/24

y = - {m x}^{2} - 2 \Rightarrow \frac{d y}{d x} = - 2 m x

\frac{d}{d x} (4 x^{2} + y^{2}) = \frac{d}{d x} (4) \Rightarrow 8 x + 2 y \times \frac{d y}{d x} = 0

\Rightarrow \frac{d y}{d x} = \frac{- 4 x}{y}

- 2 m x = \frac{- 4 x}{y}; x \neq 0 \Rightarrow m = \frac{2}{y}

[y = 0 \Rightarrow n o u n i q u e (x, y) \in R^{2}]

[y = 0 \Rightarrow x = \underline{+} 1 \Rightarrow 0 = - m - 2 \Rightarrow m = - 2]

[x = 0 \Rightarrow y = - 2]

A s f a r a s w e w a n t (x, y) \in R^{2}, t h e r e w o u l d a l w a y s

b e u n i q u e s o l u t i o n s i f y \neq 0

H o w e v e r, m = 0 p r o d u c e s u n i q u e (x, y) \in C^{2} .

Commented by efronzo1 last updated on 29/May/24

sir y = - {mx}^{2} - 2

Commented by A5T last updated on 29/May/24

S a m e t h i n g, m c o u l d b e n e g a t i v e o r p o s i t i v e

o r 0 .

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