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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

#### 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

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

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

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

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

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

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

$\text { Given that }$

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

$\Rightarrow \frac{3}{2\sqrt{3 x_1 - 2}} = 2$

$\Rightarrow 3 = 4\sqrt{3 x_1 - 2}$

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

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

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

$\Rightarrow x_1 = \frac{41}{48}$

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

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

$\text { Equation oftangentis },$

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

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

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

$\Rightarrow 24y - 18 = 48x - 41$

$\Rightarrow 48x - 24y - 23 = 0$

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#### Video TutorialsVIEW ALL [3]

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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