Question

The equation of the tangent to the curve $y=\sqrt{9-2x^2}$ as the point where the ordinate and the abscissa are equal , is

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

Answer 1

The correct option is D $2x+y-3\sqrt{3}=0$

Given that $x_1=y_1$
Therefore, $x_1=\sqrt{9-2x_1^2}$
$\Rightarrow x_1^2=9-2x_1^2$
$\Rightarrow x_1=\pm \sqrt{3}$
since, $y_1>0$, therefore the point is $(\sqrt{3},\sqrt{3})$
Also, $y=\sqrt{9-2x^2}$
$\Rightarrow y^2=9-2x^2$
On differentiating w.r.t $x$, we get
$2y\frac{dy}{dx}=-4x$
$\Rightarrow \frac{dy}{dx}=-\frac{2x}{y}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(\sqrt{3},\sqrt{3})}=-2$
So, the required equation of tangent is
$(y-\sqrt{3})=-2(x-\sqrt{3})$
$\Rightarrow 2x+y-3\sqrt{3}=0$