Question

For which of the following curves, the line $x+\sqrt{3}y=2\sqrt{3}$ is the tangent at the point $(\frac{3\sqrt{3}}{2},\frac{1}{2})$?

• A. $x^2+9y^2=9$
• B. $2x^2-18y^2=9$
• C. $y^2=\frac{x}{6\sqrt{3}}$
• D. $x^2+y^2=7$

Answer 1

The correct option is A

$x^2+9y^2=9$

Explanation for the correct options:

Step 1. Finding the equation of tangent:

The equation of a tangent to circle $x^2+y^2=a^2$ at point $(x_1,y_1)$ is : $\mathit{x}{\mathit{x}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}\mathit{y}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{}{\mathit{a}}^{\mathbf{2}}$.

From option $\left(a\right)$ we can write equation of tangent as,

$x^2+9y^2=9$

$\Rightarrow$$x{x}_1+9y{y}_1=9$

Step 2. Put the values $(\frac{3\sqrt{3}}{2},\frac{1}{2})$ in the equation, we get

$x\left(\frac{3\sqrt{3}}{2}\right)+9y\left(\frac{1}{2}\right)=9$

$\Rightarrow$ $3\sqrt{3}x+9y=18$

$\Rightarrow$ $x+\sqrt{3}y=2\sqrt{3}$

$\therefore$$x+\sqrt{3}y=2\sqrt{3}$ is tangent to the curve ${\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{9}{\mathit{y}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{9}$

Hence, Option(A) is correct