First slide

Tangents and normals

Question

The shortest distance between the line $x - y = 1$ and the curve $x^{2} = 2 y$ is

Moderate

A

$\frac{1}{2}$
B

$\frac{1}{\sqrt{2}}$
C

0
D

$\frac{1}{2 \sqrt{2}}$

Solution

Consider the curve $x^{2} = 2 y$
Differentiate both sides with respect to $x$
$\begin{matrix} 2 x = 2 \frac{d y}{d x} \\ \frac{d y}{d x} = x \end{matrix}$
slope of the given line x-y=1 is 1
therefore x=1
Point on the curve $P (1, \frac{1}{2})$ ( since x=1 $\Rightarrow$ 2y=1 )
Eq of tangent to the curve at P is
$y - \frac{1}{2} = 1 (x - 1), y = x - \frac{1}{2} \Rightarrow x - y = \frac{1}{2}$
Shortest distance = distance between the parallel lines x-y=1 & x-y= $\frac{1}{2}$ is $= |\frac{1 - \frac{1}{2}}{\sqrt{1 + 1}}| = \frac{1}{2 \sqrt{2}}$

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