The length of the latus-rectum of the parabola y2 - 8x minus- 2y - 17 - 0 is-a- 2-b- 4-c- 8-d- 16

(c) 8Given:y2 + 8x minus; 2y + 17 = 0 #8658;y 12 1+8x+17=0 #8658;y 12+8x+16=0 #8658;y 12= 8x+2Let X=x+2, #160;Y=y 1 there4; Y2= 8XComparing with y2 ...

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Standard XII

Mathematics

Property 6

The length of...

Question

The length of the latus-rectum of the parabola y² + 8x − 2y + 17 = 0 is
(a) 2
(b) 4
(c) 8
(d) 16

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Solution

(c) 8

Given:
y² + 8x − 2y + 17 = 0
$\Rightarrow {(y - 1)}^{2} - 1 + 8 x + 17 = 0 \Rightarrow {(y - 1)}^{2} + 8 x + 16 = 0 \Rightarrow {(y - 1)}^{2} = - 8 (x + 2)$

Let $X = x + 2, Y = y - 1$

∴ $Y^{2} = - 8 X$

Comparing with y²=4ax:
a = 2

Length of the latus rectum = 4a = 8 units

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The length of the latus-rectum of the parabola y2 + 8x − 2y + 17 = 0 is (a) 2 (b) 4 (c) 8 (d) 16

(c) 8Given:y2 + 8x − 2y + 17 = 0⇒y-12-1+8x+17=0⇒y-12+8x+16=0⇒y-12=-8x+2Let X=x+2, Y=y-1∴ Y2=-8XComparing with y2=4ax:a = 2Length of the latus rectum = 4a = 8 units

The length of the latus-rectum of the parabola y² + 8x − 2y + 17 = 0 is
(a) 2
(b) 4
(c) 8
(d) 16

(c) 8

Given:
y² + 8x − 2y + 17 = 0
$\Rightarrow {(y - 1)}^{2} - 1 + 8 x + 17 = 0 \Rightarrow {(y - 1)}^{2} + 8 x + 16 = 0 \Rightarrow {(y - 1)}^{2} = - 8 (x + 2)$

Let $X = x + 2, Y = y - 1$

∴ $Y^{2} = - 8 X$

Comparing with y²=4ax:
a = 2

Length of the latus rectum = 4a = 8 units