Question:

<p>Given equation of curve is<br /> y² = x - 2 ..... (i)<br /> and the equation of line is<br /> <img alt="" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/3jeqNdC.png" style="height:194px; width:250px" /><br /> y = x ..... (ii)<br /> Consider a point P(t² + 2, t) on parabola (i).<br /> For the shortest distance between curve (i) and line (ii), the line PM should be perpendicular to line (ii) and parabola (i), i.e. tangent at P should be parallel to y = x.<br /> <span class="math-tex">\(\left.\therefore \frac{d y}{d x}\right|_{\text {at point } P}\)</span>&nbsp;= Slope of tangent at point P to curve (i) [<span class="math-tex">\(\because\)</span>&nbsp;tangent is parallel to line y = x]<br /> <span class="math-tex">\(\left.\Rightarrow \quad \frac{1}{2 y}\right|_{P}=1\)</span>&nbsp;[differentiating the curve (i), we get&nbsp;<span class="math-tex">\(2 y \frac{d y}{d x}=1\)</span>]<br /> <span class="math-tex">\(\Rightarrow \frac{1}{2 t}=1 \Rightarrow t=\frac{1}{2}\)</span>&nbsp;<span class="math-tex">\(\left[\because P(x, y)=P\left(t^{2}+2, t\right)\right]\)</span><br /> So, the point P is&nbsp;<span class="math-tex">\(\left(\frac{9}{4}, \frac{1}{2}\right)\)</span><br /> Now, minimum distance = PM&nbsp;<span class="math-tex">\(=\frac{\left|\frac{9}{4}-\frac{1}{2}\right|}{\sqrt{2}}\)</span><br /> [<span class="math-tex">\(\because\)</span>&nbsp;distance of a point P(x₁, y₁) from a line ax + by + c = 0 is&nbsp;<span class="math-tex">\(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)</span>]<br /> <span class="math-tex">\(=\frac{7}{4 \sqrt{2}} \text { units }\)</span></p>