AP EAMCET · Maths · Application of Derivatives
The point on the curve \(y=x^2+4 x+3\) which is closest to the line \(y=3 x+2\) is
- A \(\left(\frac{1}{2}, \frac{5}{4}\right)\)
- B \(\left(\frac{-1}{2}, \frac{5}{4}\right)\)
- C \(\left(2, \frac{-5}{3}\right)\)
- D \(\left(2, \frac{5}{3}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{-1}{2}, \frac{5}{4}\right)\)
Step-by-step Solution
Detailed explanation
Let \((x, y)\) be on the parabola \(y=x^2+4 x+3\) which is closet to the line \(y=3 x+2\) Perpendicular distance between a point \(\left(x_1, y_1\right)\) and a line \((a x+b y+c=0)\) \(D=\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|\) \(\left[\therefore\right.\) line…
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