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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equation of the tangent to the curve \(y = \sqrt {3x - 2}\) which is parallel to the line \(4x - 2y + 5 = 0 .\)

$\begin{array}{1 1} (A)\;48x+24y=23 \\(B)\;48x-24y=23 \\ (C)\;48x-24y=24 \\(D)\;18x-24y=24 \end{array} $

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Let the point of contact of the tangent line parallel to the given line be $P(x_1,y_1)$.
The equation of the curve is $y=\sqrt{3x-2}$
Now differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}$
Hence the slope of the tangent at $(x_1,y_1)$ is
$\large\frac{dy}{dx}_{(x_1,y_1)}=\frac{3}{2\sqrt{3x_1-2}}$
Step 2:
Since the tangent is at $(x_1,y_1)$ is parallel to the line $4x-2y+5=0$
Therefore $\large\frac{dy}{dx}_{(x_1,y_1)}$$=$slope of the line $4x-2y+5=0$
Slope of the line is $-\bigg(\large\frac{coeff\;of\;x}{coeff\;of\;y}\bigg)$
$\Rightarrow -\big(\large\frac{-4}{2}\big)$$=2$
Therefore $\large\frac{3}{2\sqrt{3x_1-2}}$$=2$
$\Rightarrow 3=4\sqrt{3x_1-2}$
Step 3:
Squaring on both sides we get,
$9=16(3x_1-2)$
Therefore $3x_1=\large\frac{9}{16}$$+2$
$\qquad\qquad\quad=\large\frac{41}{16}$
Therefore $x_1=\large\frac{41}{48}$
Since $(x_1,y_1)$ lies on $y=\sqrt{3x-2}$,
$y_1=\sqrt{3x_1-2}$
$\quad=\sqrt{3\times \large\frac{41}{48}-\normalsize 2}$
$\Rightarrow y_1=\large\frac{3}{4}$
Hence the point of contact is $\big(\large\frac{41}{48},\frac{3}{4}\big)$
Step 4:
Hence the required equation of the tangent is $y-\large\frac{3}{4}$$=2(x-\large\frac{41}{48})$
$\Rightarrow \large\frac{4y-3}{4}=\frac{48x-41}{24}$
$\Rightarrow 6(4y-3)=48x-41$
Therefore $48x-24y=23$
This is the required equation of the tangent.
answered Jul 12, 2013 by sreemathi.v
 

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