The focus and directrix of a parabola are 1-1 and x-y-3-0- Its vertex isA- - 7 1- 4 4 -B- - 1- 2 -7C- - 1- 2 -7D- - 1-5- 2-2 -

The correct option is C S=(1_1, 1) ex Directrix x+y+3=0 h 1 1=k+1 1= (1 1+3) 1+1= 3 2ex h= 1 2 k= 5 2 Z=( 1 2, 5 2) and A=Mid pt of SZ=(1 4, 7 4) ...

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Standard XII

Mathematics

Standard Equation of Parabola

The focus and...

Question

The focus and directrix of a parabola are (1,-1) and x+y+3=0. Its vertex is

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Solution

The correct option is C

$S = (1_{1}, - 1) D i r e c t r i x x + y + 3 = 0$
$\frac{h - 1}{1} = \frac{k + 1}{1} = \frac{- (1 - 1 + 3)}{1 + 1} = \frac{- 3}{2} h = \frac{- 1}{2} k = \frac{- 5}{2}$
$Z = (\frac{- 1}{2}, \frac{- 5}{2})$ and A=Mid pt of SZ= $(\frac{1}{4}, \frac{- 7}{4})$

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The focus and directrix of a parabola are (1,-1) and x+y+3=0. Its vertex is

The correct option is C S=(11,−1)Directrix x+y+3=0 h−11=k+11=−(1−1+3)1+1=−32h=−12 k=−52 Z=(−12,−52) and A=Mid pt of SZ=(14,−74)

The correct option is C

$S = (1_{1}, - 1) D i r e c t r i x x + y + 3 = 0$
$\frac{h - 1}{1} = \frac{k + 1}{1} = \frac{- (1 - 1 + 3)}{1 + 1} = \frac{- 3}{2} h = \frac{- 1}{2} k = \frac{- 5}{2}$
$Z = (\frac{- 1}{2}, \frac{- 5}{2})$ and A=Mid pt of SZ= $(\frac{1}{4}, \frac{- 7}{4})$