The length of the chord intercepted by the parabola $y^{2} = 4 x$ on the straight line $x + y = 1$ is

4

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4 \sqrt{2}

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8

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8 \sqrt{2}

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Solution

The correct option is A $8$

Substitute $y = 1 - x$ in equation of parabola to get

$\Rightarrow$ $x^{2} - 6 x + 1 = 0 ⟹ x = 3 \pm 2 \sqrt{2}$

Substitute values of $x$ in equation of parabola to get intersection points

$(3 + 2 \sqrt{2}, - 2 - 2 \sqrt{2})$ and $(3 - 2 \sqrt{2}, - 2 + 2 \sqrt{2})$

$\Rightarrow$ Hence, length of intercept $= \sqrt{(3 + 2 \sqrt{2} - 3 + 2 \sqrt{2})^{2} + (- 2 + 2 \sqrt{2} + 2 + 2 \sqrt{2})^{2}} = \sqrt{32 + 32} = 8$

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Similar questions

Whats the length of chord intercepted by the parabola $y^{2} = 4 x o n t h e l i n e x + y = 0$

x^{2} + y^{2} - 8 x - 2 y - 8 = 0

x + y + 1 = 0

y^{2} = 8 x

4 x^{2} + 9 y^{2} = 1

9 y = 1

x^{2} + y^{2} = r^{2}

\frac{x}{a} + \frac{y}{b} = 1

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