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If the latus rectum of parabola $\mathrm{(y-k)^2=8(x-h)}$ always lies between the lines $\mathrm{x+y=1}$ and $\mathrm{x+y=9}$ , then $\mathrm{h+k}$ lies in the interval :
Option: 1
$[3,6]$

Option: 2
$[4,7]$

Option: 3
$[4,6]$

Option: 4
$(4,6)$

Answers (1)

best_answer

$\mathrm{(y-k)^2=4(x-h)}$ is a parabola with latus rectum parallel to y-axis.

Focus $\mathrm{S \equiv(\mathrm{h}+1, \mathrm{k})}$

Hence coordinate $\mathrm{ A \equiv(h+3, k)}$ and $\mathrm{ B \equiv(h- 1, \mathrm{k})}$

A lie on same side of origin to $\mathrm{x+y-9=0 \Rightarrow \mathrm{h}+3+\mathrm{k}-9<0 \Rightarrow \mathrm{h}+\mathrm{k}<6}$

B lie on opposite side of origin to $\mathrm{x+y-1= 0}$

$\mathrm{\mathrm{h}-1+\mathrm{k}-1>2 \Rightarrow \mathrm{h}+\mathrm{k}<4}$

Hence $\mathrm{4<\mathrm{h}+\mathrm{k}<6.}$

Posted by

SANGALDEEP SINGH

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