Find the equation of the normal to the curve x^2 + 2y^2-4x-6y+8= 0 at…

The correct answer is Option (3) → $x=2$

The given curve is $x^2 +2y^2-4x-6y+8=0$   ...(i)

Putting $x = 2$ in (i), we get

$4+2y^2-8-6y+8=0⇒2y^2-6y+4=0$

$⇒y^2-3y+2=0⇒(y-1) (y-2) = 0⇒y=1,2$.

Hence, the feet of normals to the curve (i) are (2, 1) and (2, 2).

Diff. (i) w.r.t. x, we get

$2x + 2.2y\frac{dy}{dx}-4-6\frac{dy}{dx}=0⇒(2y-3) \frac{dy}{dx}=2-x$

$\frac{dy}{dx}=-\frac{x-2}{2y-3}$   …(ii)

Normal at (2, 1)

Slope of tangent at (2, 1) = $-\frac{2-2}{2.1-3}= 0$

⇒ the tangent at (2, 1) is parallel to x-axis

⇒ the normal at (2, 1) is parallel to y-axis i.e. it is a vertical line.

∴ The equation of the normal at (2, 1) is $x = 2$.

Normal at (2, 2)

Slope of tangent at (2, 2) = $-\frac{2-2}{2.2-3}= 0$

⇒ the tangent at (2, 2) is parallel to x-axis

⇒ the normal at (2, 2) is parallel to y-axis.

∴ The equation of the normal at (2, 2) is $x = 2$.

we note that the normals at the points (2, 1) and (2, 2) to the given curve coincide i.e. the curve has same normal.