Question

The equation of the tangent to the curve $y=\sqrt{9-2x^2}$ at the point, where the ordinate & 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 $2x+y-3\sqrt{3}=0$

Let $(x_1,y_1)$ be any point on the curve such that $x_1=y_1$

$\therefore {x_1}^2=9-2x_1^2$

$\Rightarrow x_1=\pm \sqrt{3}$

$\Rightarrow y_1=\pm \sqrt{3}$

Since, $y>0$, therefore the point is $(\sqrt{3},\sqrt{3})$

We have $y^2=9-2x^2$

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{y}$

$\therefore {\left(\frac{dy}{dx}\right)}_{(\sqrt{3},\sqrt{3})}=-2$

So, the equation of tangent is

$(y-\sqrt{3})=-2(x-\sqrt{3})$

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