The point of intersection of the normals to the parabola $y^{2} = 4 x$ at the ends of its latus rectum is :

(0, 3)

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(2, 0)

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(3, 0)

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(0, 2)

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Solution

The correct option is A (0, 3)
Given that,
$y^{2} = 4 x$
end latus rectum is $(0, 3)$ .
Option $(A)$ is correct answer.,

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Similar questions

y^{2} = 4 x

y^{2} = 4 x

(x - 3)^{2} + (y + 2)^{2} = r^{2}

r^{2}

(y + 3)^{2} = 8 (x - 2)

If the normals of the parabola y²=4x drawn at the end points of its latus rectum are tangents to the circle (x-3)²+(y+2)²=r², then the value of r² is

(y + 3)^{2} = 8 (x - 2)

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