TS EAMCET · Maths · Application of Derivatives
If the point \(\mathrm{P}\left(x_1, y_1\right)\) lying on the curve \(y=x^2-x+1\) is the closest point to the line \(y=x-3\) then the perpendicular distance from P to the line \(3 x+4 y-2=0\) is
- A \(\frac{16}{5}\)
- B 4
- C 1
- D \(\frac{7}{5}\)
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
Tangent slope \(m_{tan} = 2x-1\). Line slope \(m_{line} = 1\). \(2x-1 = 1 \Rightarrow 2x=2 \Rightarrow x_1=1\). \(y_1 = (1)^2-1+1 = 1\). \(\mathrm{P}(1,1)\). Distance \(D = \frac{|3(1)+4(1)-2|}{\sqrt{3^2+4^2}}\). \(D = \frac{|3+4-2|}{\sqrt{9+16}}\).…
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