Question

Find the equation of tangent to the ellipse $3x^2+4y^2=12$ which are parallel to the line $y=x-1$

Answer 1

Equation of ellipse is

$\frac{x^2}{2^2}+\frac{y^2}{{\sqrt{3}}^2}=1$

where $a^2=4b^2=3$

Slope of the line $y=x-1$ is $1$

Now equation of tangent havig slope $m$ is

$y=mx+\sqrt{a^2{m}^2+b^2}$

If the tangent is parallel to the given line , then $m=1$

Equation of the tangent is $y=x-1$