Hyperbola (1 Viewer)

[xwrathbringerx]

nd the acute angle between the tangents which can be drawn from (2,3) to the hyperbolas x^2 - y^2 = 2. I've found dy/dx = x/y but I don't know what to do with it. Do I simply sub in (2,3)? If so, what do I do next???<!-- google_ad_section_end -->

 

[Drongoski]

nd the acute angle between the tangents which can be drawn from (2,3) to the hyperbolas x^2 - y^2 = 2. I've found dy/dx = x/y but I don't know what to do with it. Do I simply sub in (2,3)? If so, what do I do next???<!-- google_ad_section_end -->

1 way is: you know the chord of contact of the 2 tangents to the hyperbola from point (2,3) is given by the formula: xx₁ - yy₁ = 2

You then find the the 2 points of contact of these 2 tangents (1 way is to simultaneosly solve for the eqn of the hyperbola and the chord of contact: 2x - 3y = 2 [this is just the eqn of a straight line]. You can find the gradients of the 2 tangents and thence the tangent of the angle between them[ tan (theta) = (m1-m2+/(1+m1 x m2) ; take absolute value of this if this value is negative [in general, when 2 lines intersect, there are 2 angles of intersection: the acute one and the obtuse one. tan of the acute angle is positive and of the obtuse, negative]. This then yields the desired angle.