Partial differentiation and tangency $\text{Q}$ Write the equation of tangent at the vertex of the parabola $2y^2+3y+4x-3=0$ . $\text{My attempt}$ if I partially differentiate the curve with y then I get $y+0.75=0$ which is correct answer. now i am not much in calculus but I know that $dy/dx$ is the equation of tangent but here I have used partial differentiation . Am I wrong ?and is there more strong geometrical approach?
 A: Your approach will work but it will also make things way more complicated.  No calculus required.
Express $x$ in terms of $y$ and get:
$$x = -\frac{1}{2}y^2 -\frac{3}{4}y + \frac{3}{4}$$
This is a parabola that opens to the left.  How do we know this?  It's $x$ as a function of $y$, which means it opens either to the left or to the right.  And the coefficient on $y^2$ is negative, so it opens to the left.
The line tangent to the parabola at the vertex is, in this case, the vertical line through the vertex (vertical because the parabola opens to the left [as opposed to opening up or down]).  Therefore the equation will have the form $x = h$, where $h$ is the $x$-coordinate of the vertex.  So let's find that.
Now, just like with parabolas that are $y$ as a function of $x$, we can find the vertex of this parabola by analyzing the coefficients.  We have $a = -1/2$, $b = -3/4$, and $c = 3/4$.  The vertex is the point
$$ \left(g\left(-\frac{b}{2a}\right), -\frac{b}{2a}\right),$$
where $\displaystyle g(y) = x = -\frac{1}{2}y^2 -\frac{3}{4}y + \frac{3}{4}$.
Can you take it from here?
A: If you take partial derivatives, $(\partial f/\partial x, \partial f/\partial y)=(u,v)$, you will get a gradient, i.e., a normal vector to the curve in point $(x,y)$. From it, you can recover a tangent vector by taking $(v,-u)$ which has a null dot profuct with the gradient.
But it is much more direct to use the rule
$$2y^2+3y+4x-3=0 \rightarrow 2yy_0+3 \frac12(y+y_0)+4 \frac12(x+x_0)-3=0$$
which gives you at once the equation of the tangent in $(x_0,y_0)$.
Have you seen this rule with its different cases:
$$x^2 \rightarrow xx_0, \ y^2 \rightarrow yy_0, xy \rightarrow \frac12(xx_0+yy_0), \ x \rightarrow \frac12(x+x_0),  \ y \rightarrow \frac12(y+y_0), \ constant \ c \rightarrow c $$
