Consider the parabola 25[(x-2)^2+(y+5)^2]=(3x | vedclass

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(A) The given equation is $25[(x-2)^2+(y+5)^2]=(3x+4y-1)^2$. This is of the form $SP^2 = e^2 PM^2$,where $S(2, -5)$ is the focus and $3x+4y-1=0$ is th

Vedclass · Accepted Answer

$I-B, II-E, III-C, IV-D$

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(A) The given equation is $25[(x-2)^2+(y+5)^2]=(3x+4y-1)^2$. This is of the form $SP^2 = e^2 PM^2$,where $S(2, -5)$ is the focus and $3x+4y-1=0$ is the directrix,with $e=1$.$I$. Vertex: The vertex is the midpoint of the focus $S(2, -5)$ and the projection of the focus on the directrix. The projection of $S(2, -5)$ on $3x+4y-1=0$ is $P' = (x, y)$ such that $\frac{x-2}{3} = \frac{y+5}{4} = -\frac{3(2)+4(-5)-1}{3^2+4^2} = -\frac{-15}{25} = \frac{3}{5}$. Thus,$x = 2 + \frac{9}{5} = \frac{19}{5}$ and $y = -5 + \frac{12}{5} = -\frac{13}{5}$. The vertex is the midpoint of $S(2, -5)$ and $(\frac{19}{5}, -\frac{13}{5})$,which is $(\frac{2+19/5}{2}, \frac{-5-13/5}{2}) = (\frac{29}{10}, \frac{-38}{10})$. Thus,$I-B$.$II$. Length of latus rectum: The distance from the focus $(2, -5)$ to the directrix $3x+4y-1=0$ is $d = \frac{|3(2)+4(-5)-1|}{\sqrt{3^2+4^2}} = \frac{|6-20-1|}{5} = \frac{15}{5} = 3$. The length of the latus rectum is $2d = 2 	imes 3 = 6$. Thus,$II-E$.$III$. Directrix: Given as $3x+4y-1=0$. Thus,$III-C$.$IV$. One end of the latus rectum: The latus rectum is a line through the focus $(2, -5)$ parallel to the directrix $3x+4y-1=0$,i.e.,$3x+4y+k=0$. Since it passes through $(2, -5)$,$3(2)+4(-5)+k=0 \Rightarrow k=14$. So,$3x+4y+14=0$. The intersection of this line with the parabola gives the ends of the latus rectum,which are $(\frac{-2}{5}, \frac{-16}{5})$ and $(\frac{22}{5}, \frac{-34}{5})$. Thus,$IV-D$.

List-$I$	List-$II$
$I$. Vertex	$A$. $8$
$II$. Length of latus rectum	$B$. $(\frac{29}{10}, \frac{-38}{10})$
$III$. Directrix	$C$. $3x+4y-1=0$
$IV$. One end of the latus rectum	$D$. $(\frac{-2}{5}, \frac{-16}{5})$
	$E$. $6$

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