1. prove that the equaion of parabola whose axis of symmetry is parallel to y axisis given as (x−h)^2 =+_− 4p(y−k) 2. if Dis a perpendicular distance of the point p(r,t)from the line (L):(x/r)+(y/t)=1 then find the value of D why not helped me??


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Question Number 182514 by Best1 last updated on 11/Dec/22

$1 . p r o v e t h a t t h e e q u a i o n o f p a r a b o l a$ $w h o s e a x i s o f s y m m e t r y i s p a r a l l e l$ $t o y a x i s i s g i v e n a s {(x - h)}^{2} = \underline{+} 4 p (y - k)$ $2 . i f D i s a p e r p e n d i c u l a r d i s t a n c e$ $o f t h e p o i n t p (r, t) f r o m t h e l i n e$ $(L) : \frac{x}{r} + \frac{y}{t} = 1 t h e n f i n d t h e$ $v a l u e o f D$ $w h y n o t h e l p e d m e ? ?$
Commented by mr W last updated on 11/Dec/22

$d o y o u m e a n p e o p l e m u s t h e l p y o u,$ $n o m a t t e r i f t h e y c a n o r {c a n}^{'} t a n d i f$ $t h e y w i l l o r {w o n}^{'} t ? a n d t h e y s h o u l d$ $t e l l w h y i f t h e y {d o n}^{'} t ({c a n}^{'} t o r {w o n}^{'} t)$ $h e l p y o u ?$
	Answered by mr W last updated on 11/Dec/22

	$2)$ $d i s t a n c e f r o m p o i n t (u, v) t o l i n e$ $a x + b x + c = 0 i s d = \frac{∣ a u + b v + c ∣}{\sqrt{a^{2} + b^{2}}} .$ $a p p l y i n g t h i s y o u c a n g e t$ $D = \frac{∣ \frac{r}{r} + \frac{t}{t} - 1 ∣}{\sqrt{\frac{1}{r^{2}} + \frac{1}{t^{2}}}} = \frac{∣ r t ∣}{\sqrt{r^{2} + t^{2}}}$
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