Find the length of the chord of the parabola $y^{2} = 8 x$ , whose equation is $x + y = 1$ .

8 \sqrt{3}

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4 \sqrt{3}

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2 \sqrt{3}

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\sqrt{3}

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Solution

The correct option is A $8 \sqrt{3}$
Substitute $x = 1 - y$ in $y^{2} = 8 x$
Then $y^{2} = 8 (1 - y)$
$y = \frac{8 \pm \sqrt{96}}{2} = 4 \pm 2 \sqrt{6}$
Therefore, $x = 3 \mp 2 \sqrt{6}$
Using distance formula length of chord $= 8 \sqrt{3}$

Suggest Corrections

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x^{2} = 4 y

x - \sqrt{2} y + 4 \sqrt{2} = 0

List-l	List - II
A. Length of the latusrectum of the parabola $3 y^{2} = - 4 x$	1. $(5, 2)$
B.Equation of the parabola whose focus $($ 0,4 $)$ and directrix $x + 4 = 0$ is	2. $\frac{- 4}{3}$
C. The focus of the parabola $(y - 2)^{2} = 8 (x - 3)$ is	3. $\frac{4}{3}$
D. Equation of the parabola whose vertex $(- 1, - 2)$ latusrectum 4 and axis is parallel to y-axis	4. $(x + 1)^{2} = 4 (y + 2)$
	5. $x^{2} = 16 y$