Question

Equation of the tangent of $3x^2+4y^2=12$ parallel to $x-2y+1=0$ is:

• A. $x-2y+7=0$
• B. $x-2y+4=0$
• C. $x-2y+5=0$
• D. $x-2y+9=0$

Answer 1

The correct option is B $x-2y+4=0$

Equation of line parallel to $x-2y+1=0$ is

given by, $x-2y+c=0$

Now given ellipse is, $3x^2+4y^2=12$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1$

comparing it with standard ellipse $a^2=4,b^2=3$

Now using condition of tangency, $c^2=a^2{m}^2+b^2$

$\Rightarrow c^2=4\times \left(\frac{1}{4}\right)+3=4$ [$\because$ slope of line (1) is $\frac{1}{2}]$

$\Rightarrow c=\pm 2$

Hence required tangent is, $x-2y\pm 4=0$