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Question Number 135816 by ShiimiGee last updated on 16/Mar/21

Commented by ShiimiGee last updated on 16/Mar/21

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Commented by liberty last updated on 16/Mar/21

Q1  (a)ℓ=(√((3−0)^2 +(4−0)^2 )) = 5  (b) ℓ=(√((1+2)^2 +(1+3)^2 )) = 5  (c) ℓ=(√((−1+4)^2 +(−2+6)^2 )) = 5  (d) ℓ=(√((2+3)^2 +(4−16)^2 )) = 13  (e) ℓ=(√((11+1)^2 +(−2−3)^2 )) = 13

$${Q}\mathrm{1} \\ $$$$\left({a}\right)\ell=\sqrt{\left(\mathrm{3}−\mathrm{0}\right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{0}\right)^{\mathrm{2}} }\:=\:\mathrm{5} \\ $$$$\left({b}\right)\:\ell=\sqrt{\left(\mathrm{1}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{3}\right)^{\mathrm{2}} }\:=\:\mathrm{5} \\ $$$$\left({c}\right)\:\ell=\sqrt{\left(−\mathrm{1}+\mathrm{4}\right)^{\mathrm{2}} +\left(−\mathrm{2}+\mathrm{6}\right)^{\mathrm{2}} }\:=\:\mathrm{5} \\ $$$$\left({d}\right)\:\ell=\sqrt{\left(\mathrm{2}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{16}\right)^{\mathrm{2}} }\:=\:\mathrm{13} \\ $$$$\left({e}\right)\:\ell=\sqrt{\left(\mathrm{11}+\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{2}−\mathrm{3}\right)^{\mathrm{2}} }\:=\:\mathrm{13} \\ $$

Commented by EDWIN88 last updated on 16/Mar/21

Q2.  (i)  { ((x=rcos α)),((y=rsin α)) :} ⇒x^2 +y^2  = r^2    ∴ x^2 +y^2 −2y=0 ⇒r^2 −2r sin α = 0  (ii) r^2  sin^2 α = 4a(a−r cos α)      r^2  sin^2 α +4ar cos α = 4a^2

$$\mathrm{Q2}. \\ $$$$\left(\mathrm{i}\right)\:\begin{cases}{\mathrm{x}=\mathrm{rcos}\:\alpha}\\{\mathrm{y}=\mathrm{rsin}\:\alpha}\end{cases}\:\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{r}^{\mathrm{2}} \\ $$$$\:\therefore\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2y}=\mathrm{0}\:\Rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{2r}\:\mathrm{sin}\:\alpha\:=\:\mathrm{0} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{r}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} \alpha\:=\:\mathrm{4}{a}\left({a}−{r}\:\mathrm{cos}\:\alpha\right) \\ $$$$\:\:\:\:\mathrm{r}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} \alpha\:+\mathrm{4}{ar}\:\mathrm{cos}\:\alpha\:=\:\mathrm{4}{a}^{\mathrm{2}} \\ $$

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