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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of tangent to the ellipse $3x^2+4y^2=12$ which are parallel to the line $y+2x=4$

$\begin{array}{1 1}(A)\;2x+19 \\(B)\;-2x+\sqrt {19} \\(C)\;2x-\sqrt {19} \\(D)\;x+\sqrt {19} \end{array}$

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1 Answer

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Equation of ellipse is
$\large\frac{x^2}{4}+\frac{y^2}{3}$$=1$
$a^2=4\;b^2=3$
Slope of the line $y+2x=4$ is $-2$
Now equation of a tangent having slope m is
$y= mx +\sqrt {a^2m^2+b^2}$
If the tangent is parallel to the given line , then
$m=-2$
Equation of the tangent is $y= -2x +\sqrt {19}$
Hence B is the correct answer.
answered Apr 10, 2014 by meena.p
 

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