AM-1781

Tangents and normals

A Math · Secondary 4 · Hard · Tangents and normals

The curve has equation

y = \sqrt{3x+4}

. Find the equation of the normal to the curve at the point where

x = 2

.

$y-\sqrt{10} = -\dfrac{2\sqrt{10}}{3}(x-2)$
$y-\sqrt{10} = \dfrac{2\sqrt{10}}{3}(x-2)$
$y-\sqrt{10} = -\dfrac{3}{2\sqrt{10}}(x-2)$
$y-2 = -\dfrac{2\sqrt{10}}{3}(x-\sqrt{10})$

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