Question

Find the coordinates of a point of the parabola y = x² + 7x + 2 which is closest to the straight line y = 3x − 3.

Answer 1

Let the coordinates of the required point of the parabola y = x² + 7x + 2 be (h, k).

\[\therefore k = h^2 + 7h + 2\]   ..... (1)

Distance of the point (h, k) from the straight line y = 3x − 3 is given by \[D = \frac{\left| 3h - k - 3 \right|}{\sqrt{\left( 3 \right)^2 + \left( - 1 \right)^2}} = \frac{\left| 3h - k - 3 \right|}{\sqrt{10}}\]  .......(2)

From (1) and (2), we get

\[D = \frac{\left| 3h - h^2 - 7h - 2 - 3 \right|}{\sqrt{10}}\]

\[= \frac{\left| - h^2 - 4h - 5 \right|}{\sqrt{10}}\]

\[= \frac{h^2 + 4h + 5}{\sqrt{10}}\]

Differentiating both sides w.r.t. h, we get

\[\frac{dD}{dh} = \frac{2h + 4}{\sqrt{10}}\]

For maxima and minima,

\[\frac{dD}{dh} = 0\]

\[ \Rightarrow \frac{2h + 4}{\sqrt{10}} = 0\]

\[ \Rightarrow 2h = - 4\]

\[ \Rightarrow h = - 2\]

Now,

\[\frac{d^2 D}{d^2 h} = \frac{2}{\sqrt{10}} > 0\]

So, h = −2 is the point of minima.
When h = −2,

\[k = \left( - 2 \right)^2 + 7 \times \left( - 2 \right) + 2 = 4 - 14 + 2 = - 8\]   [Using (1)]

Thus, the coordinates of a point of the given parabola that is closest to the given straight line is (−2, −8).