Question:
The closest point on the parabola
\[
y=x^2+7x+2
\]
to the straight line
\[
y=3x-2
\]
is
The closest point on the parabola
\[
y=x^2+7x+2
\]
to the straight line
\[
y=3x-2
\]
is
Show Hint
To find the point on a curve nearest to a line, substitute the curve equation into the line-distance formula and minimize the resulting expression. Often it simplifies into a perfect square.
Updated On: Jul 9, 2026
- \((-1,-4)\)
- \((1,10)\)
- \((-2,-8)\)
\((0,2)\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
The point on a curve nearest to a line minimizes the perpendicular distance from the point to the line.
The given line is
\[
3x-y-2=0.
\]
For a point \((x,y)\) on the parabola
\[
y=x^2+7x+2,
\]
the perpendicular distance is proportional to
\[
|3x-y-2|.
\]
Hence we minimize its square.
Step 1: Express the distance in terms of \(x\). Substitute \[ y=x^2+7x+2. \] Then \[ 3x-y-2 = 3x-(x^2+7x+2)-2. \] \[ =-(x^2+4x+4). \] \[ =-(x+2)^2. \] Thus \[ |3x-y-2| = (x+2)^2. \] The distance from the line is \[ d = \frac{(x+2)^2}{\sqrt{10}}. \]
Step 2: Minimize the distance. Since \[ d=\frac{(x+2)^2}{\sqrt{10}}, \] the minimum value occurs when \[ (x+2)^2=0. \] \[ x=-2. \]
Step 3: Find the corresponding point on the parabola. Substitute \(x=-2\) into \[ y=x^2+7x+2. \] \[ y=4-14+2. \] \[ y=-8. \] Hence the required point is \[ (-2,-8). \]
Step 4: Write the final answer. \[ \boxed{(-2,-8)} \]
Step 1: Express the distance in terms of \(x\). Substitute \[ y=x^2+7x+2. \] Then \[ 3x-y-2 = 3x-(x^2+7x+2)-2. \] \[ =-(x^2+4x+4). \] \[ =-(x+2)^2. \] Thus \[ |3x-y-2| = (x+2)^2. \] The distance from the line is \[ d = \frac{(x+2)^2}{\sqrt{10}}. \]
Step 2: Minimize the distance. Since \[ d=\frac{(x+2)^2}{\sqrt{10}}, \] the minimum value occurs when \[ (x+2)^2=0. \] \[ x=-2. \]
Step 3: Find the corresponding point on the parabola. Substitute \(x=-2\) into \[ y=x^2+7x+2. \] \[ y=4-14+2. \] \[ y=-8. \] Hence the required point is \[ (-2,-8). \]
Step 4: Write the final answer. \[ \boxed{(-2,-8)} \]
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