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At What Points on the Curve Y = X2 − 4x + 5 is the Tangent Perpendicular to the Line 2y + X = 7? - Mathematics

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Question

At what points on the curve y = x2 − 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

Solution

Let (x1y1) 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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Chapter 16: Tangents and Normals - Exercise 16.1 [Page 11]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 16 Tangents and Normals
Exercise 16.1 | Q 16 | Page 11

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