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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the equation of the tangent to the curve \( x^2+3y=3\), which is parallel to the line $ y-4x+5=0$

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  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
Step 1:
Given curve is $x^2+3y=3$
Let $y=\large\frac{-x^2+3}{3}$
On differentiating with respect to $x$
$\therefore \large\frac{dy}{dx}=\frac{1}{3}$$[-2x]$
$\qquad=\large\frac{-2}{3}$$x$
Step 2:
Since the tangent is parallel to the line $y-4x+5=0$
Then slopes should be equal
Slope of the given line is 4
$\large\frac{-2x}{3}=$$4$
$x=\large\frac{-4\times 3}{2}$
$\;\;=\large\frac{-12}{2}$
$\;\;=-6$
Step 3:
$\therefore y=\large\frac{-(-6)^2+3}{3}$
$\;\;\quad=\large\frac{-36+3}{3}$
$\;\;\quad=\large\frac{-33}{3}$
$\;\;\quad=-11$
Hence the points of contact are $(-6,-11)$
Step 4:
Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
(i.e)$[y-(-11)]=4(x-(-6))$
$y+11=4x+6$
$\Rightarrow 4x-y=11-6$
$\Rightarrow 4x-y=5$
Hence $4x-y=5$ is the required equation.
answered Oct 1, 2013 by sreemathi.v
 
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