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Solution for Find the Equation of the Tangent to the Curve Y = √ 3 X − 2 Which is Parallel to the 4x − 2y + 5 = 0 ? - CBSE (Commerce) Class 12 - Mathematics

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Question

Find the equation of the tangent to the curve  \[y = \sqrt{3x - 2}\] which is parallel to the 4x − 2y + 5 = 0 ?

Solution

Slope of the given line is 2

\[\text { Let }\left( x_1 , y_1 \right)\text { be the point where the tangent is drawn to the curvey }= \sqrt{3x - 2}\]

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence }, y_1 = \sqrt{3 x_1 - 2} . . . \left( 1 \right)\]

\[\text { Now }, y = \sqrt{3x - 2}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{3}{2\sqrt{3x - 2}}\]

\[\text { Slope of tangent at} \left( x_1 , y_1 \right) =\frac{3}{2\sqrt{3 x_1 - 2}}\]

\[\text { Given that }\]

\[\text { Slope oftangent= slope of the given line }\]

\[ \Rightarrow \frac{3}{2\sqrt{3 x_1 - 2}} = 2\]

\[ \Rightarrow 3 = 4\sqrt{3 x_1 - 2}\]

\[ \Rightarrow 9 = 16\left( 3 x_1 - 2 \right)\]

\[ \Rightarrow \frac{9}{16} = 3 x_1 - 2\]

\[ \Rightarrow 3 x_1 = \frac{9}{16} + 2 = \frac{9 + 32}{16} = \frac{41}{16}\]

\[ \Rightarrow x_1 = \frac{41}{48}\]

\[\text { Now,} y_1 = \sqrt{\frac{123}{48} - 2} = \sqrt{\frac{27}{48}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \left[ \text { From }(1) \right]\]

\[ \therefore \left( x_1 , y_1 \right) = \left( \frac{41}{48}, \frac{3}{4} \right)\]

\[\text { Equation oftangentis },\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{3}{4} = 2 \left( x - \frac{41}{48} \right)\]

\[ \Rightarrow \frac{4y - 3}{4} = 2\left( \frac{48x - 41}{48} \right)\]

\[ \Rightarrow 24y - 18 = 48x - 41\]

\[ \Rightarrow 48x - 24y - 23 = 0\]

  Is there an error in this question or solution?
Solution for question: Find the Equation of the Tangent to the Curve Y = √ 3 X − 2 Which is Parallel to the 4x − 2y + 5 = 0 ? concept: Tangents and Normals. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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