Question:
The point on the curve
\[
y=x^2+4x+3
\]
which is closest to the line
\[
y=3x+2
\]
is
The point on the curve
\[
y=x^2+4x+3
\]
which is closest to the line
\[
y=3x+2
\]
is
Show Hint
To find the point on a curve closest to a line, write a general point on the curve and minimize its perpendicular distance from the line.
Updated On: Jun 22, 2026
- \(\left(\frac{1}{2},\frac{5}{4}\right)\)
- \(\left(-\frac{1}{2},\frac{5}{4}\right)\)
- \(\left(2,-\frac{5}{3}\right)\)
- \(\left(2,\frac{5}{3}\right)\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write the general point on the curve.
Given curve is \[ y=x^2+4x+3 \] So, any point on the curve can be written as \[ P(x,\ x^2+4x+3) \]
Step 2: Write the given line in standard form.
Given line is \[ y=3x+2 \] So, \[ 3x-y+2=0 \]
Step 3: Use distance formula from point to line.
Distance of point \(P(x,\ x^2+4x+3)\) from the line \[ 3x-y+2=0 \] is \[ d=\frac{|3x-(x^2+4x+3)+2|}{\sqrt{3^2+(-1)^2}} \] \[ d=\frac{|-x^2-x-1|}{\sqrt{10}} \] Since \[ x^2+x+1\gt 0 \] for all real \(x\), we get \[ d=\frac{x^2+x+1}{\sqrt{10}} \]
Step 4: Minimize the distance.
To minimize \(d\), we minimize \[ x^2+x+1 \] This quadratic has minimum at \[ x=-\frac{b}{2a} \] Here, \[ a=1,\qquad b=1 \] So, \[ x=-\frac{1}{2} \]
Step 5: Find the corresponding value of \(y\).
Substitute \[ x=-\frac{1}{2} \] in \[ y=x^2+4x+3 \] \[ y=\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+3 \] \[ y=\frac{1}{4}-2+3 \] \[ y=\frac{1}{4}+1 \] \[ y=\frac{5}{4} \]
Step 6: Final conclusion.
Therefore, the required point is \[ \boxed{\left(-\frac{1}{2},\frac{5}{4}\right)} \]
Given curve is \[ y=x^2+4x+3 \] So, any point on the curve can be written as \[ P(x,\ x^2+4x+3) \]
Step 2: Write the given line in standard form.
Given line is \[ y=3x+2 \] So, \[ 3x-y+2=0 \]
Step 3: Use distance formula from point to line.
Distance of point \(P(x,\ x^2+4x+3)\) from the line \[ 3x-y+2=0 \] is \[ d=\frac{|3x-(x^2+4x+3)+2|}{\sqrt{3^2+(-1)^2}} \] \[ d=\frac{|-x^2-x-1|}{\sqrt{10}} \] Since \[ x^2+x+1\gt 0 \] for all real \(x\), we get \[ d=\frac{x^2+x+1}{\sqrt{10}} \]
Step 4: Minimize the distance.
To minimize \(d\), we minimize \[ x^2+x+1 \] This quadratic has minimum at \[ x=-\frac{b}{2a} \] Here, \[ a=1,\qquad b=1 \] So, \[ x=-\frac{1}{2} \]
Step 5: Find the corresponding value of \(y\).
Substitute \[ x=-\frac{1}{2} \] in \[ y=x^2+4x+3 \] \[ y=\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+3 \] \[ y=\frac{1}{4}-2+3 \] \[ y=\frac{1}{4}+1 \] \[ y=\frac{5}{4} \]
Step 6: Final conclusion.
Therefore, the required point is \[ \boxed{\left(-\frac{1}{2},\frac{5}{4}\right)} \]
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