Question

Equation of tangent to the curve $y=\sqrt{9−2x^2}$ at the point where ordinate and abscissa are equal, is

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

Answer 1

Answer: (D)

Given $y=\sqrt{9−2x^2}$
Put $y=x$, we get
$x=\sqrt{9−2x^2}⇒x^2=9−2x^2\text{ (squaring on both sides) }$
$⇒3x^2−9⇒x^2=3$
$∴x=Â\pm \sqrt{3}$
So, $(x=\sqrt{3},y=\sqrt{3})$ and $(x=−\sqrt{3},y=−\sqrt{3})$ (reject)
Now, ${\frac{dy}{dx}]}_{(\sqrt{3},\sqrt{3})}=\frac{(−4x)}{2\sqrt{9−2x^2}}=\frac{−2x}{\sqrt{9−2x^2}}=\frac{−2\sqrt{3}}{\sqrt{3}}=−2$
So, equation of tangent is $(y−\sqrt{3})=−2(x−\sqrt{3})⇒2x+y=3\sqrt{3}$