Finding tangent of a point x=d on general ellipse formula I was wondering if I could have some help with calculus.
I had a question asked, find the equation of the tangent at the point where $x=3/2$ on ellipse $4x^2+9y^2=36$.
I have an answer:
Ellipse equation, $y=-\frac{2\sqrt3}9x+\frac{4\sqrt3}3$
and the radial equation is $y=-\frac{2\sqrt3}3x$
That was all well and good, however, now I must find the equation of the tangent at point where $x=d$ on the general ellipse $ax^2+bx^2=c$ (where $x$ is in the first quadrant). Please help me, as my tangent equation is never ending and the calculus is becoming mixed for me. If you could include all steps, I would be grateful.
Thankyou
 A: The equation of the ellipse is
$$ax^2+by^2=c$$
Differentiating with respect to $x$,
$$2ax+2byy'=0$$
$$y'=-\frac{ax}{by}$$
The tangent to a curve at a point has a slope equal to the slope of the curve at that point. So, the slope of the tangent at $x=d$ will be $y'$ evaluated at $x=d$.
At $x=d$,
$$ad^2+by^2=c$$
$$y=\pm\frac{c-ad^2}{b}$$
Let this value be labelled $Y$. The appropriate $y$ based on the fact that the point is in the first quadrant must be taken.
Now, the general equation of the tangent at $(d,Y)$ is given by,
$$y-Y=y'(d,Y)(x-d)$$
which can be obtained by substituting the respective values.
A: One form of equation for a line through a point $\mathbf p$ is $\mathbf n\cdot(\mathbf x-\mathbf p)=0$, where $\mathbf n$ is a vector perpendicular (normal) to the line. The normal to a curve at a point is, of course, perpendicular to the tangent line at that point.  
For a curve given by $f(x,y)=c$, the gradient $\nabla f=\left({\partial f\over\partial x},{\partial f\over\partial y}\right)$ is always normal to the curve. Thus, we have in general that an equation for the tangent line at a point $\mathbf p$ on this curve is $\nabla f(\mathbf p)\cdot(\mathbf x-\mathbf p)=0$, or, letting $\mathbf p=(x_0,y_0)$,$$f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)=0.$$ In this case, $f(x,y) = ax^2+by^2$, so an equation of the tangent at the point $(x_0,y_0)$ on the ellipse is $ax_0(x-x_0)+by_0(y-y_0)=0$ (I’ve dropped the common factor of $2$ that came from $\nabla f$). Plug in your values for $x_0$ and $y_0$ and rearrange to suit.
