Let $$M$$ and $$N$$ be the number of points on the curve $$y^5 - 9xy + 2x = 0$$, where the tangents to the curve are parallel to $$x$$-axis and $$y$$-axis, respectively. Then the value of $$M + N$$ equals

Given

$$y^5-9xy+2x=0$$

Rearrange:

$$x(2-9y)+y^5=0$$

$$x=\frac{y^5}{9y-2}$$

Differentiate implicitly:

$$5y^4-9\left(x+y\frac{dx}{dy}\right)+2\frac{dx}{dy}=0$$

or directly differentiate

$$x=\frac{y^5}{9y-2}$$

Using the quotient rule,

$$\frac{dx}{dy}=\frac{5y^4(9y-2)-9y^5}{(9y-2)^2}$$

$$=\frac{y^4(36y-10)}{(9y-2)^2}$$

### Tangents parallel to the x-axis

A tangent parallel to the

$$x$$

-axis means

$$\frac{dy}{dx}=0$$

which is equivalent to

$$\frac{dx}{dy}\neq0$$

and undefined reciprocal, so we set

$$\frac{dx}{dy}=\infty$$

This occurs when

$$(9y-2)^2=0$$

$$y=\frac29$$

Substituting into the curve,

$$x=\frac{(2/9)^5}{9(2/9)-2}$$

gives denominator

$$0$$

which is impossible.

Hence there is no point with

$$9y-2=0$$

on the curve.

Therefore we instead use

$$\frac{dy}{dx}=0$$

from implicit differentiation.

Differentiating:

$$5y^4y'-9(xy'+y)+2=0$$

$$(5y^4-9x)y'=9y-2$$

Thus

$$y'=\frac{9y-2}{5y^4-9x}$$

For tangents parallel to the

$$x$$

-axis,

$$y'=0$$

$$9y-2=0$$

$$y=\frac29$$

Substituting into the curve:

$$\left(\frac29\right)^5-9x\left(\frac29\right)+2x=0$$

The coefficient of

$$x$$

vanishes, leaving

$$\left(\frac29\right)^5=0$$

which is impossible.

Hence

$$M=0$$

### Tangents parallel to the y-axis

A tangent parallel to the

$$y$$

-axis requires

$$5y^4-9x=0$$

and

$$9y-2\neq0$$

From

$$5y^4-9x=0$$

$$x=\frac59y^4$$

Substitute into the curve:

$$y^5-9\left(\frac59y^4\right)y+2\left(\frac59y^4\right)=0$$

$$y^5-5y^5+\frac{10}{9}y^4=0$$

$$y^4\left(-4y+\frac{10}{9}\right)=0$$

Hence

$$y=0$$

or

$$y=\frac{5}{18}$$

These give two distinct points on the curve, and for both,

$$9y-2\neq0$$

Therefore

$$N=2$$

Thus

$$M+N=0+2$$

$$=2$$

Final Answer :

$$2$$