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Q.
Equation of tangent to the curve $y=\sqrt{9-2 x^2}$ at the point where ordinate and abscissa are equal, is
Application of Derivatives
Solution:
Given $y =\sqrt{9-2 x ^2}$
Put $y=x$, we get
$x =\sqrt{9-2 x ^2} \Rightarrow x ^2=9-2 x ^2 \text { (squaring on both sides) } $
$\Rightarrow 3 x ^2-9 \Rightarrow x ^2=3$
$\therefore x = \pm \sqrt{3}$
So, $( x =\sqrt{3}, y =\sqrt{3})$ and $( x =-\sqrt{3}, y =-\sqrt{3}) $ (reject)
Now, $\left.\frac{ dy }{ dx }\right]_{(\sqrt{3}, \sqrt{3})}=\frac{(-4 x )}{2 \sqrt{9-2 x ^2}}=\frac{-2 x }{\sqrt{9-2 x ^2}}=\frac{-2 \sqrt{3}}{\sqrt{3}}=-2$
So, equation of tangent is $(y-\sqrt{3})=-2(x-\sqrt{3}) \Rightarrow 2 x+y=3 \sqrt{3}$