Tangents to Hyperbolas
In this section, we discuss tangents in different forms for hyperbolas. We also discuss chords of contact, chords bisected at given points, pair of tangents from external points and so on, as we have already done for the case of other conic sections. The forms of the equations for all these will turn out to be the same as the results obtained for other conics.
We will use the hyperbola x2a2−y2b2=1x2a2−y2b2=1 in our discussion.
TANGENT AT P (x1, y1) : Consider a point P(x1,y1)P(x1,y1) on the given hyperbola. The slope of the tangent at this point can be obtained by differentiating the equation of the hyperbola.
x2a2−y2b2=1⇒2xa2−2yb2dydx=0⇒dydx=b2xa2y⇒dydx|P(x1,y1)=b2x1a2y1x2a2−y2b2=1⇒2xa2−2yb2dydx=0⇒dydx=b2xa2y⇒dydx∣∣∣P(x1,y1)=b2x1a2y1
Thus, the equation of the tangent will be
y−y1=b2x1a2y1(x−x1)⇒xx1a2−yy1b2=x21a2−y21b2⇒xx1a2−yy1b2=1(RHS above is 1 since (x1,y1)lies on the hyperbola)y−y1=b2x1a2y1(x−x1)⇒xx1a2−yy1b2=x21a2−y21b2⇒xx1a2−yy1b2=1(RHS above is 1 since (x1,y1)lies on the hyperbola)
This equation is sometimes written more concisely as T(x1,y1)=0T(x1,y1)=0
TANGENT AT P (a secθθ, b tanθθ ) : If the point P has been specified in parametric form, the equation for the tangent at P can be obtained by the substitution x1→asecθx1→asecθ and y1→btanθy1→btanθ in the equation for the tangent we obtained above :
xasecθ−ybtanθ=1xasecθ−ybtanθ=1
This is the equation of the tangent at the point θ.θ.
TANGENT OF SLOPE m : In Example -10 in the previous section, we’ve already obtained the equation of any tangent of slope m to the hyperbola x2a2−y2b2=1:x2a2−y2b2=1:
y=mx±√a2m2−b2y=mx±√a2m2−b2
The coordinates of the point(s) of contact, as already mentioned earlier, are
(±a2m√a2m2−b2,±b2√a2m2−b2)(±a2m√a2m2−b2,±b2√a2m2−b2)
Example - 11
Show that two tangents can be drawn to a hyperbola from any point P lying outside the parabola.
Solution : Let the equation of the hyperbola be x2a2−y2b2=1x2a2−y2b2=1 and the coordinates of P be (h, k).
Any tangent of slope m to this hyperbola will have the equation
y=mx±√a2m2−b2y=mx±√a2m2−b2
If this passes through P(h,k),P(h,k), we have
k=mh±√a2m2−b2⇒(k−mh)2=a2m2−b2⇒(h2−a2)m2−2hkm+k2+b2=0k=mh±√a2m2−b2⇒(k−mh)2=a2m2−b2⇒(h2−a2)m2−2hkm+k2+b2=0
This is a quadratic in m which will give us two roots, corresponding to two tangents. These roots are real if the discriminant is positive, i.e., if
4h2k2>4(h2−a2)(k2+b2)⇒h2b2<a2k2+a2b2⇒h2a2−k2b2<1⇒S(h,k)<04h2k2>4(h2−a2)(k2+b2)⇒h2b2<a2k2+a2b2⇒h2a2−k2b2<1⇒S(h,k)<0
Thus, two real tangents to the hyperbola can be drawn from any point only if that point lies outside the hyperbola. If a point lies on a hyperbola, only one tangent can be drawn (basically, coincident tangents) at the same point, while for any point inside the hyperbola, tangents cannot be drawn to the hyperbola.
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