Question

Find the equation of the tangent to x²/a² -y²/b² = 1 at (x₁,y₁)

Answer 1

Our hyperbola has equation x^2/a^2−y^2/b^2=1 We want dy/dx. Use implicit differentiation. We get 2x/a^2−(2y/b^2)(dydx)=0............(1) Now we can find dydxdydx at our point (x1,y1). That gives the slope of the tangent line, and now we can find the equation of the tangent line. The slope of the normal is −1/m where m is the slope of the tangent. So now we know the slope of the normal. It passes through (x1,y1), so we can find its equation. Remark: For the normal, there is a minor complication if x1=±a (and therefore y1=0). In those two cases, the normal is vertical. Added: There seems to be difficulty in finding the equation of the tangent line. From (1) we have dy/dx=(2xb^2)/(2ya^2).So at the point (x1,y1)on the hyperbola, the slope of the tangent line is equal to (x1*b^2)/(y1*a^2). It follows that the equation of the tangent line is (y−y1)/(x−x1)=(x1*b^2)/(y1*a^2). Now we do some algebraic manipulation. Multiply through by (x−x1)(y1*a^2). We get (y−y1)y1*a^2=(x−x1)(x1*b^2). This simplifies to x*x1*b^2−y*y1*a^2=(x1^2)b^2−(y1^2)a^2. Divide through by a^2b^2, and use the fact that (x1)^2/a^2−(y1)^2/b^2=1,We end up with (x.x1)/a^2−(y.y1)/b^2=1 Note :this the best proof to find the tangent to a hyperbola