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प्रश्न
At what points on the curve y = x2 − 4x + 5 is the tangent perpendicular to the line 2y + x = 7?
उत्तर
Let (x1, y1) be the required point.
Slope of the given line = \[\frac{- 1}{2}\]
Slope of the line perpendicular to this line = 2
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence }, y_1 = {x_1}^2 - 4 x_1 + 5 . . . \left( 1 \right)\]
\[\text { Now }, y = x^2 - 4x + 5 \]
\[ \therefore \frac{dy}{dx} = 2x - 4\]
\[\text { Now }, \]
\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =2 x_1 -4\]
\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)=\text { Slope of the given line [Given]}\]
\[ \therefore 2 x_1 - 4 = 2\]
\[ \Rightarrow 2 x_1 = 6\]
\[ \Rightarrow x_1 = 3\]
\[\text {Also }, \]
\[ y_1 = 9 - 12 + 5 = 2 [\text { From eq. } (1)]\]
\[\text { Thus, the required point is }\left( 3, 2 \right).\]
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