Question

Find the equation of the tangent to the ellipse $x^2+2y^2=4$ at the point where ordinate is 1

• A. $x+\sqrt{2y}-2\sqrt{2}=0\&-x+\sqrt{2y}-2\sqrt{2}$
• B. $x-\sqrt{2y}-2\sqrt{2}=0\&x-\sqrt{2y}+2\sqrt{2}$
• C. $x+\sqrt{2y}+2\sqrt{2}=0\&x+\sqrt{2y}+2\sqrt{2}$
• D. none of these

Answer 1

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

$x^2+2y^2=4y=1,x=\pm \sqrt{2}$

Differentiating w.r.t $x$

$2x+4y\frac{dy}{dx}=0\frac{dy}{dx}=-\frac{x}{2y}\frac{dy}{dx}|y=1,x=\sqrt{2}=-\frac{1}{\sqrt{2}}\frac{dy}{dx}|y=1,x=-\sqrt{2}=\frac{1}{\sqrt{2}}$

Equation

$y-1=\frac{-1}{\sqrt{2}}(x-\sqrt{2})y-1=\frac{1}{\sqrt{2}}(x+\sqrt{2})$

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