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Question Number 90347 by mr W last updated on 22/Apr/20

Commented by mr W last updated on 22/Apr/20

$t h e p a r a b o l a h a s t h e s a m e s h a p e a s$ $y = x^{2} . f i n d {i t}^{'} s e q u a t i o n i f i t p a s s e s$ $t h r o u g h O, A, B a s s h o w n .$ $[f o r s o m e r e a s o n a j f o u r s i r h a s d e l e t e d$ $t h i s q u e s t i o n . b u t i t i s a g o o d q u e s t i o n .]$
Commented by MJS last updated on 23/Apr/20
$let a=p and b=q to not get confused ax^2 +bxy+cy^2 +dx+ey+f=0 D=ac−(b^2 /4) this is a { ((ellipse; D>0)),((parabola; D=0)),((hyperbola; D<0)) :} ⇒ (1) ac−(b^2 /4)=0 ((0),(0) ) ∈par ⇒ (2) f=0 ((p),(0) ) ∈par ⇒ (3) ap^2 +dp+f=0 ((0),(q) ) ∈par ⇒ (4) cq^2 +eq+f=0 (5) we know the shape is similar to y=x^2 the angle of rotation: tan 2θ =(b/(a−c)); if b=c ⇒ ⇒ θ=45° usually we rotate like this: ((x),(y) ) = (((x′cos θ −y′sin θ)),((x′sin θ +y′cos θ)) ) and get ax^2 +bxy+cy^2 +dx+ey+f=0 → → Ax′^2 +Cy′^2 +Dx′+Ey′+F and then complete the squares to see the translation in x and y directions here we have (y′−n)=(x′−m)^2 → → x′^2 −2mx′−y′+(m^2 −n)=0 ⇒ A=1; C=0; D=−2m; E=−1; F=m^2 −n now we must find θ I stop here and go to bed. this is what′s already there: (1) ac−(b^2 /4)=0 (2) f=0 (3) ap+d=0 (4) cq+e=0$
$let a = p and b = q to not get confused$ ${a x}^{2} + b x y + {c y}^{2} + d x + e y + f = 0$ $D = a c - \frac{b^{2}}{4}$ $this is a {\begin{cases} ellipse; D > 0 \\ parabola; D = 0 \\ hyperbola; D < 0 \end{cases}$ $\Rightarrow (1) a c - \frac{b^{2}}{4} = 0$ $(\begin{matrix} 0 \\ 0 \end{matrix}) \in par \Rightarrow (2) f = 0$ $(\begin{matrix} p \\ 0 \end{matrix}) \in par \Rightarrow (3) {a p}^{2} + d p + f = 0$ $(\begin{matrix} 0 \\ q \end{matrix}) \in par \Rightarrow (4) {c q}^{2} + e q + f = 0$ $(5) we know the shape is similar to y = x^{2}$ $the angle of rotation : \tan 2 θ = \frac{b}{a - c}; if b = c \Rightarrow$ $\Rightarrow θ = 45 °$ $usually we rotate like this : (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x^{'} \cos θ - y^{'} \sin θ \\ x^{'} \sin θ + y^{'} \cos θ \end{matrix})$ $and get$ ${a x}^{2} + b x y + {c y}^{2} + d x + e y + f = 0 \to$ $\to {A x}^{' 2} + {C y}^{' 2} + {D x}^{'} + {E y}^{'} + F$ $and then complete the squares to see the$ $translation in x and y directions$ $here we have (y^{'} - n) = {(x^{'} - m)}^{2} \to$ $\to x^{' 2} - 2 {m x}^{'} - y^{'} + (m^{2} - n) = 0$ $\Rightarrow A = 1; C = 0; D = - 2 m; E = - 1; F = m^{2} - n$ $now we must find θ$ $I stop here and go to bed . this is {what}^{'} s$ $already there :$ $(1) a c - \frac{b^{2}}{4} = 0$ $(2) f = 0$ $(3) a p + d = 0$ $(4) c q + e = 0$
Commented by TawaTawa1 last updated on 23/Apr/20

$Sir, i have come again . please help me with : Q 90362$
Commented by TawaTawa1 last updated on 23/Apr/20

$Please . God bless you more for everytime .$
Commented by mr W last updated on 23/Apr/20

$t h a n k y o u M J S s i r!$
	Answered by ajfour last updated on 23/Apr/20

	Commented by ajfour last updated on 23/Apr/20

	$\frac{y - m x - c}{\sqrt{1 + m^{2}}} = \sqrt{{(x - h)}^{2} + {(y - k)}^{2}}$ $O (0, 0), A (a, 0), B (0, b) l i e o n$ $p a r a b o l a .$ $\Rightarrow h^{2} + k^{2} = \frac{c^{2}}{1 + m^{2}} . . . . (i)$ ${(a - h)}^{2} + k^{2} = \frac{{(- a m - c)}^{2}}{1 + m^{2}} . . (i i)$ $h^{2} + {(b - k)}^{2} = \frac{{(b - c)}^{2}}{1 + m^{2}} . . . (i i i)$ $A n d s i n c e t h i s p a r a b o l a h a s$ $s a m e s h a p e a s x^{2} = 4 (\frac{1}{4}) y$ $\frac{k - m h - c}{\sqrt{1 + m^{2}}} = \frac{1}{2} . . . . (i v)$ $A g a i n u s i n g (i) i n (i i)$ $a^{2} - 2 a h = \frac{a^{2} m^{2} + 2 a m c}{1 + m^{2}}$ $h = \frac{\frac{a}{2} - c m}{1 + m^{2}} . . . . . . (I)$ $i n (i i i) :$ $b^{2} - 2 b k = \frac{b^{2} - 2 b c}{1 + m^{2}}$ $k = \frac{\frac{{b m}^{2}}{2} + c}{1 + m^{2}} . . . . . (I I)$ $a n d \frac{- c}{\sqrt{1 + m^{2}}} = \sqrt{h^{2} + k^{2}}$ $= \frac{1}{2} + \frac{m h - k}{\sqrt{1 + m^{2}}}$ $\frac{c^{2}}{1 + m^{2}} = {(\frac{\frac{a}{2} - c m}{1 + m^{2}})}^{2} + {(\frac{\frac{{b m}^{2}}{2} + c}{1 + m^{2}})}^{2}$ $\Rightarrow \frac{a^{2}}{4} - a c m + \frac{b^{2} m^{4}}{4} + {b c m}^{2} = 0$ $c = \frac{a^{2} + b^{2} m^{4}}{4 m (a - b m)} . . . . . . . (I I I)$ $a n d$ $\frac{\frac{{b m}^{2}}{2} + c}{1 + m^{2}} - m (\frac{\frac{a}{2} - c m}{1 + m^{2}}) - c = \frac{\sqrt{1 + m^{2}}}{2}$ $\Rightarrow {b m}^{2} - a m = {(1 + m^{2})}^{3 / 2} . . . . (I V)$ $h e r e f r o m w e g e t m .$ $t h e n c, a n d t h e n h, k u s i n g e q s .$ $(I I I), (I I), a n d (I) .$ $E q . o f p a r a b o l a i s$ $\frac{y - m x - c}{\sqrt{1 + m^{2}}} = \sqrt{{(x - h)}^{2} + {(y - k)}^{2}}$ $◼$
	Commented by mr W last updated on 23/Apr/20

	$t h a n k s a l o t s i r! i^{'} l l a t t e m p t a s i m i l a r$ $w a y .$
	Commented by ajfour last updated on 23/Apr/20

	$F o r a = \sqrt{3}, b = 5 / 3 w e g e t$ $m = - \sqrt{3}, c = - \frac{7}{8}, h = - \frac{3 \sqrt{3}}{32},$ $a n d k = \frac{13}{32} .$ $A l s o m \approx - 0.56775$ $a n d c o r r e s p o n d i n g v a l u e s o f$ $c, h, k .$
	Commented by ajfour last updated on 23/Apr/20

	Answered by mr W last updated on 23/Apr/20

	Commented by mr W last updated on 23/Apr/20

	$s i n c e t h e p a r a b o l a h a s t h e s a m e s h a p e$ $a s y = x^{2}, i t c a n b e o b t a i n e d f r o m y = x^{2}$ $t h r o u g h s o m e t r a n s f o r m a t i o n :$ $t r a n s l a t i o n : v e r t e x f r o m O (0, 0) t o O^{'} (- h, - h^{2})$ $y = x^{2} \Rightarrow y = x^{2} - 2 h x$ $r o t a t i o n : a b o u t O b y a n g l e θ$ $\Rightarrow x \sin θ + y \cos θ = {(x \cos θ - y \sin θ)}^{2} - 2 h (x \cos θ - y \sin θ)$ $s i n c e t h i s t r a n s f o r m e d p a r a b o l a$ $p a s s e s t h r o u g h (a, 0) a n d (0, b), w e g e t :$ $b \cos θ = {(- b \sin θ)}^{2} - 2 h (- b \sin θ)$ $\cos θ = b \sin^{2} θ + 2 h \sin θ$ $\Rightarrow h = \frac{1}{2} (\frac{1}{\tan θ} - b \sin θ) . . . (i)$ $a \sin θ = {(a \cos θ)}^{2} - 2 h (a \cos θ)$ $\sin θ = a \cos^{2} θ - 2 h \cos θ$ $\Rightarrow h = \frac{1}{2} (a \cos θ - \tan θ) . . . (i i)$ $\frac{1}{\tan θ} - b \sin θ = a \cos θ - \tan θ$ $\Rightarrow \frac{1}{\tan θ} + \tan θ = a \cos θ + b \sin θ . . (i i i)$ $e q n . o f p a r a b o l a :$ $x \sin θ + y \cos θ = {(x \cos θ - y \sin θ)}^{2} - (a \cos θ - \tan θ) (x \cos θ - y \sin θ)$ $u s u a l l y (i i i) h a s t w o r o o t s f o r θ,$ $t h a t m e a n s w e h a v e u s u a l l y t w o$ $p o s s i b l e p a r a b o l a s, s e e e x a m p l e s :$
	Commented by mr W last updated on 23/Apr/20

	Commented by mr W last updated on 23/Apr/20

	Commented by mr W last updated on 23/Apr/20

	Commented by ajfour last updated on 23/Apr/20

	$I t s v e r y n i c e, s h o r t a n d w i s e$ $w a y S i r; t h a n k s a l o t .$
	Commented by mr W last updated on 23/Apr/20

	Commented by mr W last updated on 23/Apr/20

	$i n g e n e r a l a r i g h t - a n g l e d t r i a n g l e$ $c a n b e i n s c r i b e d i n t h e p a r a b o l a i n$ $t w o d i f f e r e n t w a y s, t h e r e f o r e w e$ $h a v e t w o p o s s i b l e p o s i t i o n s f o r t h e$ $p a r a b o l a .$ $t h e r e i s o n l y o n e s o l u t i o n i f a = b = \sqrt{2} .$
	Commented by ajfour last updated on 24/Apr/20

	$t o o g o o d i n t e r p r e t a t i o n, a n d$ $e x p l a n a t i o n, S i r!$
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