find the tangent line to the ellipse x^2/a^2+y^2/b^2=1

Question

please help me to answer this equation in my differential calculus step by step please

Answer 1

You need a point at which to find the tangent line.

Differentiate:

2x/a²+(2y/b²)dy/dx=0.

The slope of the tangent line is dy/dx=(2x/a²)/(2y/b²)=b²x/(a²y).

To find the tangent line at a point (p,q) we first find the gradient=b²p/(a²q).

The gradient must pass through (p,q) so:

y-q=(b²p/(a²q))(x-p), y=b²px/(a²q)-b²p²/(a²q)+q,

y=(b²px-b²p²+a²q²)/(a²q).

a, b, p, q are all constants so we end up with the line y=mx+c where:

m=b²p/(a²q), c=(a²q²-b²p²)/(a²q) (which is the y-intercept).

There are 4 special tangent lines at (a,0), (-a,0), (0,b), (0,-b) which are vertical and horizontal lines:

x=a, x=-a, y=b, y=-b.