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Question Number 58400 by rahul 19 last updated on 22/Apr/19
If parabola y=−x^2 −2x+k touches  the parabola y=−(1/2)x^2 −4x+3 , then  the value of k is ?  a) 1    b)2     c)3     d)4
$${If}\:{parabola}\:{y}=−{x}^{\mathrm{2}} −\mathrm{2}{x}+{k}\:{touches} \\ $$$${the}\:{parabola}\:{y}=−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\:,\:{then} \\ $$$${the}\:{value}\:{of}\:{k}\:{is}\:? \\ $$$$\left.{a}\left.\right)\left.\:\left.\mathrm{1}\:\:\:\:{b}\right)\mathrm{2}\:\:\:\:\:{c}\right)\mathrm{3}\:\:\:\:\:{d}\right)\mathrm{4} \\ $$
Commented by rahul 19 last updated on 22/Apr/19
i′m getting (3/2) but ans. is a!.
$${i}'{m}\:{getting}\:\frac{\mathrm{3}}{\mathrm{2}}\:{but}\:{ans}.\:{is}\:{a}!. \\ $$
Answered by mr W last updated on 22/Apr/19
−x^2 −2x+k=−(1/2)x^2 −4x+3  (1/2)x^2 −2x+3−k=0  x^2 −4x+2(3−k)=0  Δ=4^2 −4×2(3−k)=0  ⇒k=1  ⇒answer (a)
$$−{x}^{\mathrm{2}} −\mathrm{2}{x}+{k}=−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}−{k}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\left(\mathrm{3}−{k}\right)=\mathrm{0} \\ $$$$\Delta=\mathrm{4}^{\mathrm{2}} −\mathrm{4}×\mathrm{2}\left(\mathrm{3}−{k}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{1} \\ $$$$\Rightarrow{answer}\:\left({a}\right) \\ $$
Commented by rahul 19 last updated on 23/Apr/19
thanks sir!
$${thanks}\:{sir}! \\ $$

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