Question

The equation of the tangent to the ellipse $4x^2+3y^2=5$ which is parallel to the straight line $y=3x+7$ is:

• A. $2\sqrt{3}x-6\sqrt{3}y+\sqrt{155}=0$
• B. $2\sqrt{3}x-2\sqrt{3}y-\sqrt{155}=0$
• C. $6\sqrt{3}x-2\sqrt{3}y+\sqrt{155}=0$
• D. $6\sqrt{3}x-2\sqrt{3}y-\sqrt{155}=0$

Answer 1

Answer: (C), (D)

$6\sqrt{3}x-2\sqrt{3}y+\sqrt{155}=0$
$6\sqrt{3}x-2\sqrt{3}y-\sqrt{155}=0$

The slope of the required tangent is $m=3$.
Equation of tangent with slope $m$ for an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $y=mx\pm \sqrt{a^2{m}^2+b^2}$
Hence, for the given ellipse the equation of tangent is $y=3x\pm$ $\sqrt{\frac{45}{4}+\frac{5}{3}}$

$(\because a^2=\frac{5}{4},b^2=\frac{5}{3})$
$\Rightarrow y=3x\pm \sqrt{\frac{155}{12}}$
i.e. $6\sqrt{3}x-2\sqrt{3}y\pm \sqrt{155}=0$