What are the changes that have to be made in tangent equation if I want to use it for ($a<b$) ellipse as well..??

While studying parabola, we had derived the equation of tangent for $y^2=4ax$, given by $y=mx+a/m$. Now if I want to use this for $x^2=4by$, I change a $\to$ b and x $\to$ y. Now suppose for the ellipse $\frac {x^2}{a^2} + \frac{y^2}{b^2} =1 (a>b)$, I know the tangent is given by $y=mx \pm \sqrt{a^2m^2+b^2}$. What should I change to get my equation for the ellipse where $(a<b)$

It would be so great if I get to know all places where changes are to be applied.

Thankyou

• Why should anything change? Also, what about the two vertical tangents? They can’t be expressed in the form $y=my+b$. – amd Feb 4 '18 at 7:11
• @amd The reason I wasn't clear on tangents was because when a<b, eccentricity changes from e^2=1-(b^2/a^2) to e^2=1-(a^2/b^2). And I agree with your point about vertical tangents. – Ishan Sharma Feb 4 '18 at 7:29
• True, the eccentricity must be computed differently, but where does an ellipse’s eccentricity enter into the computation of tangents to it? I would guess that the formula you’ve given is derived by finding values of $c$ so that $y=mx+c$ has a single intersection with the ellipse. That only requires knowing the semi-axis lengths, not their relative magnitudes. – amd Feb 4 '18 at 20:05

Your use of different parameters for the two families of parabolas is, I think, muddying the waters. Transforming the equations of tangents to one family into equations of tangents to the other really only requires a single substitution. The two families of parabolas are each other’s reflections in the line $x=y$, and the equations of one family are obtained from the other by applying this reflection—by swapping $x$ and $y$ in the equations. This same reflection applies to the equations of the tangents, so from $y=mx+a/m$ for the tangents to $y^2=4ax$ you get $x=my+a/m$ for the tangents to $x^2=4ay$. Note, however, that the geometric meaning of the parameter $m$ has changed in the process: instead of being the slope of the tangent line, it is the reciprocal of the slope of $x=my+a/m$.

For an ellipse given by ${x^2\over a^2}+{y^2\over b^2}=1$, the formula you have holds regardless of the relative magnitudes of $a$ and $b$. Recall how the equations $y=mx\pm\sqrt{m^2a^2=b^2}$ were derived: starting with the generic equation $y=mx+d$ of a (non-vertical) line, one then finds the values of $d$ for which the line has exactly one intersection with the ellipse. The solution to that problem certainly depends on the semi-axis lengths, but whether the major axis is parallel to the $x$- or $y$-axis doesn’t enter into it.

If you’re not convinced, you could proceed as with the parabolas: the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$ is the reflection of ${x^2\over b^2}+{y^2\over a^2}=1$ in the line $x=y$, so if $a\lt b$ reflect the ellipse. By your formula, tangents to this reflected ellipse will have the form $y=mx\pm\sqrt{b^2m^2+a^2}$, which becomes $x=my\pm\sqrt{b^2m^2+a^2}$ when reflected back. Solving for $y$ results in $$y=\frac xm\pm\sqrt{{a^2\over m^2}+b^2}$$ but remember that $m$ is now the reciprocal slope, so we replace it with $m'=1/m$ to get $y=m'x\pm\sqrt{a^2{m'}^2+b^2}$.

I’m not sure that it’s really worth memorizing all of these special-case formulas, even for a timed exam. If you remember how they’re derived, it doesn’t take much time to rederive them or to apply the derivations to the specific cases in the exam. On the other hand, I think that it’s well worth learning pole-polar relationships of tangents to conics. Those will let you quickly find the tangent at any point on any conic, not just the ones that are nicely aligned with the coordinate axes, and can also be used to find the tangents to a conic through an arbitrary point. As a special case of the latter, by choosing the appropriate point at infinity, one can find tangents with a given slope as your formulas do. The key ideas for this are that the tangent at a point on a conic is that point’s polar, and that the intersections of an exterior point’s polar with the conic are the points of tangency of the tangents through that exterior point.

The tangent is again given by the same equation . The condition $a>b$ has nothing to do with the equation of tangent in this case.

• Thankyou so much sir for your answer. May I ask which equations change and which remain unchanged..? – Ishan Sharma Feb 4 '18 at 7:37
• @Ishan Sharma Are you talking about the equations of tangent? – Deepak M S Feb 4 '18 at 13:49
• I am talking about equations of normal, equation of chord joining two parametric points, etc. – Ishan Sharma Feb 4 '18 at 16:20

The restraint $a>b$ has no bearing on the result required for the envelope of all conic section tangents. Better to avoid such shortcuts and go in for the basis/root that forms tangents by differentiation, the $C$ discriminant method in seeking singular solutions of a group of tangents.

Eliminate C between

$$f(x,y,C)=0, \, \frac{\partial f }{\partial C} =0$$

Above Clairaut's procedure for parabola

$$y= m x + a/m ,\quad m= \sqrt{a/x } \quad \rightarrow y^2= 4 a x$$

same Clairaut's procedure for ellipse

$$y=mx \pm \sqrt{a^2m^2+b^2}, \quad m=\pm \frac{x \, b/a}{\sqrt { a^2 -x^2 } } \quad \rightarrow (x/a)^2 + (y/b) ^2=1$$

You can go from ellipse to hyperbola and vice-versa but not parabola to ellipse and vice-versa with the same felicity .. the application/generalization is imho too far fetched.

• Thankyou so much for your answer sir. I love to go by derivatives. But I am currently preparing for an exam where I am required to learn the basic formulae for quick response (conic sections). I go by basic approach for general curve. – Ishan Sharma Feb 4 '18 at 11:55
• @Ishan Sharma Are you preparing for IIT JEE? – Deepak M S Feb 4 '18 at 13:45
• @Deepakms Yes Sir. – Ishan Sharma Feb 4 '18 at 16:18