Question

Find the equations of tangent and normal to the ellipse $2x^2+3y^2=11$ at the point whose ordinate is 1.

Answer 1

$2x^2+3y^2=11$
At $y=1$
$x=\sqrt{\frac{11-3}{2}}$
$=\sqrt{4}$
$=2$
Hence the point is $(2,1)$
Now slope of the tangent at the point of contact $(2,1)$ can be found by differentiating the above equation of ellipse
$4x+6y.y^{\prime }=0$
Or
$y^{\prime }=\frac{-2x}{3y}$
$y_{2,1}^{\prime }=\frac{-4}{3}$
Hence the equation of the tangent is
$\frac{y-1}{x-2}=\frac{-4}{3}$
Or
$3y-3=-4x+8$
$4x+3y=11$ ...(i)
Since the normal at $(2,1)$ will be perpendicular to the tangent, hence its equation will be of the form $3x-4y=C.$
Since it passes through $(2,1)$ hence $C=(3\times 2)-(4\times 1)=6-4=2$
Therefore the equation of the normal is $3x-4y=2$.