Question

Find the shortest distance of the point (0,c) from the parabola \[y={{x}^{2}}\] where \[0\le c\le 5\].
(a) \[\sqrt{\dfrac{4c-1}{2}}\]
(b) \[\sqrt{\dfrac{4c-2}{4}}\]
(c) \[\sqrt{\dfrac{4c-3}{3}}\]
(d) \[\sqrt{\dfrac{4c-1}{4}}\]

Answer 1

Hint: Draw the graph for the parabola \[y={{x}^{2}}\] with vertex at (0,0) and draw a normal from P (0,c) to the parabola. This normal will be the shortest distance between the point and the parabola. The best approach is using calculus we can find the slope of Normal.

Complete step-by-step solution:
The parabola \[y={{x}^{2}}\] is drawn below along with normal from P (0, c) to the parabola.

Let the normal intersect the parabola at \[Q\left( \alpha ,{{\alpha }^{2}} \right)\]. Hence, PQ will the shortest distance from point P to the parabola. We have to find the distance between point P and Q.
Finding the slope of normal drawn, we get,
\[y={{x}^{2}}\]
Differentiating both sides with respect to x, we get,
\[\Rightarrow \dfrac{dy}{dx}=2x\]
Now, at \[\dfrac{dy}{dx}\] at Q will give the slope of tangent,
\[\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{at\text{ }Q}}=2\alpha \]
Since the tangent and normal are perpendicular to each other, i.e., \[{{m}_{1}}{{m}_{2}}=-1\], hence, the slope of normal will be given by
\[{{m}_{1}}{{m}_{2}}=-1\]
\[\Rightarrow 2\alpha {{m}_{1}}=-1\]
$\Rightarrow {{m}_{1}}=\dfrac{-1}{2\alpha }$
Slope of Normal, \[m=-\dfrac{1}{2\alpha }...............(i)\]
Also, since we have coordinates of two points P and Q, so using the formula of two – point form to find the slope of normal.
The two – point form is given by the equation \[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}................(ii)\]
From the given coordinates P (0, c) and \[Q\left( \alpha ,{{\alpha }^{2}} \right)\], we have,
\[{{x}_{1}}=0,{{y}_{1}}=c\]
\[{{x}_{2}}=\alpha ,{{y}_{2}}={{\alpha }^{2}}\]
Now, replacing the values of \[{{x}_{1}},{{y}_{1}},{{x}_{2}},{{y}_{2}}\] in equation (i), we get,
\[m=\dfrac{{{\alpha }^{2}}-c}{\alpha -0}..............(iii)\]
From equation (i) and (iii), we get,
\[-\dfrac{1}{2\alpha }=\dfrac{{{\alpha }^{2}}-c}{\alpha -0}\]
\[\Rightarrow -\dfrac{1}{2\alpha }=\dfrac{{{\alpha }^{2}}-c}{\alpha }\]
\[\Rightarrow -\dfrac{1}{2}={{\alpha }^{2}}-c\]
\[\Rightarrow {{\alpha }^{2}}-c=-\dfrac{1}{2}\]
\[\Rightarrow {{\alpha }^{2}}=c-\dfrac{1}{2}\]
\[\Rightarrow \alpha =\pm \sqrt{c-\dfrac{1}{2}}\]
Thus, the coordinates of point Q is \[\left( \pm \sqrt{c-\dfrac{1}{2}},c-\dfrac{1}{2} \right)\].
We know, the distance between the two points is given by the formula:
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}...............(iv)\]
Using the formula (iv) to find the distance between the point PQ, we get,
\[{{x}_{1}}=0,{{y}_{1}}=c\]
\[{{x}_{2}}=\pm \sqrt{c-\dfrac{1}{2}},{{y}_{2}}=c-\dfrac{1}{2}\]
Now, replacing the values of \[{{x}_{1}},{{y}_{1}},{{x}_{2}},{{y}_{2}}\] in equation (iv), we get,
\[PQ=\sqrt{{{\left( 0\mp \sqrt{c-\dfrac{1}{2}} \right)}^{2}}+{{\left( c-\dfrac{1}{2}-c \right)}^{2}}}\]
\[\Rightarrow PQ=\sqrt{{{\left( \mp \sqrt{c-\dfrac{1}{2}} \right)}^{2}}+{{\left( c-c-\dfrac{1}{2} \right)}^{2}}}\]
\[\Rightarrow PQ=\sqrt{\left( c-\dfrac{1}{2} \right)+{{\left( -\dfrac{1}{2} \right)}^{2}}}\]
\[\Rightarrow PQ=\sqrt{c-\dfrac{1}{2}+\dfrac{1}{4}}\]
\[\Rightarrow PQ=\sqrt{c-\dfrac{1}{4}}\]
\[\therefore PQ=\sqrt{\dfrac{4c-1}{4}}\]

Note: The best approach to solve this question is to make use of calculus. In order to find the shortest distance from a point to any geometric figure is to draw a normal from the point to the figure. Students must take care of the difference between the tangent and normal. There is a relation between the slopes of a tangent and normal, i.e., \[{{m}_{1}}{{m}_{2}}=-1\].