Q.Show that the line $x-y+4=0$ is a tangent to the ellipse $x^2+3y^2=12$. Also find the coordinates of the point of contact.

A tangent touches a curve at exactly one point; the normal at that point is perpendicular to the tangent there.

Parabola $y^2=4ax$: tangent at $(x_1,y_1)$: $yy_1=2a(x+x_1)$; at parameter $t$ (point $(at^2,2at)$): $yt=x+at^2$. Normal at $(x_1,y_1)$: $xy_1+2ay=x_1y_1+2ay_1$; at $t$: $y+xt=2at+at^3$. Theorem 5.6: three normals can be drawn from any point (a cubic in the slope), at least one always real.

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$: tangent at $(x_1,y_1)$: $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$; at $\theta$: $\frac{x\cos\theta}a+\frac{y\sin\theta}b=1$. Normal at $(x_1,y_1)$: $\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$; at $\theta$: $\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}=a^2-b^2$.

Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$: tangent at $(x_1,y_1)$: $\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1$; at $\theta$: $\frac{x\sec\theta}a-\frac{y\tan\theta}b=1$. Normal at $(x_1,y_1)$: $\frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2$; at $\theta$: $ax\cos\theta+by\cot\theta=a^2+b^2$.

Tangency of $y=mx+c$ (when only a slope, not a point, is given):

Results. Two tangents always exist from an external point (to any of the three curves); four normals from an external point (to an ellipse or hyperbola). Director circle — locus of perpendicular-tangent intersections — is the directrix $x=-a$ for the parabola, $x^2+y^2=a^2+b^2$ for the ellipse, $x^2+y^2=a^2-b^2$ for the hyperbola.