Question

The equation of tangent to the ellipse $x^2+4y^2=5$ at (-1, 1), is

• A. $x+4y+5=0$
• B. $x-4y-5=0$
• C. $x+4y-5=0$
• D. $x-4y+5=0$

Answer 1

Answer: (D)

$x-4y+5=0$

Consider the given ellipse equation.

$x^2+4y^2=5$ ……. (1)

On differentiating w.r.t $x$, we get

$2x+8y\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{x}{4y}$

${\left(\frac{dy}{dx}\right)}_{(-1,1)}=\frac{1}{4}$

Therefore, equation of the tangent equation

$y-y_1=\frac{dy}{dx}(x-x_1)$

$y-1=\frac{1}{4}(x+1)$

$4y-4=x+1$

$x-4y+5=0$

Hence, this is the answer.