co-ordinate geometry- ellipse

Question

1. Consider the conic C in the Euclidean plane described by the following equation: $3x^2 + 2y^2 + 7xy + 4x + 5y + 3 = 0$

Find all points of intersection between the line x = 0 and C. Find the equations of the tangents to C at these points.

All i can do so far is equate x to 0 and i get a quadratic with y, giving two sets of points $P1(0,-3/2)$ and $P2(0,-1)$. To get the gradient i implicitly differentiated with respect to x, isolating $dy/dy$ to one side getting $dy/dx=6x+4/-7y$ subbing in values from $P1$ and $P2$ i get $M1=8/21$ and $M2=4/7$ respectively. Shouldn't one of the gradient be negavtive? am i going the right way and what am i to do next?

Answer 1

$x = 0\\ 2y^2 + 5y + 3 = 0\\ (2y+3)(y+1) = 0$

$\frac {d}{dx}( 3x^2 + 2y^2 + 7xy + 4x+5y + 3 =0)\\ (6x + 7y + 4) + (4y + 7x + 5) y' = 0\\ y' = -\frac {(6x + 7y + 4)}{ (4y + 7x + 5)}$

And evaluate at $(0,-1)$ and $(0,-\frac 32)$

Answer 2

To find the tangent to any curve from any point lying on it, use the basic eqn

Let the point be (x1,y1)

Use, T=0

T=axx1+byy1+2h((xy1+yx1)/2)+g((x+x1)/2)+f((y+y1)/2)+c

Here, one point is (0,-1), hence the tangent to this point is

3x(0) + 2y(-1) + 7((-x+y(0))/2) + 4((x+0)/2) + 5((y-1)/2) + 3=0

** solve this and you will get the answer
y-3x+1=0** similarly solve for the other point (0,-3/2)