Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $C$ be a curve that is given by the equation:

$$ 2x^2 + y^2 = 1 $$

and let P be a point $(1,1)$, which lies outside of the curve.

We want to find all lines that are tangent to $C$ and intersect $P$, and have found $y=1$, but are not sure how to find the other line.

share|improve this question
3  
Write the general equation of a line that contains $P$. It should have one parameter $m$. Substitute in the equation of $C$ to get the intersection equation. Tangency means that this intersection equation (quadratic) has a double solution. This means its discriminant is $0$. This leads to a quadratic equation in the parameter $m$. You should find two solutions which is consistent with the geometric intuition. –  marwalix Aug 17 '11 at 13:43

4 Answers 4

up vote 2 down vote accepted

Let the equation of the line be $y = mx+c$. Since the line passes through $(1,1)$, we have $1 = m + c$ i.e. $c = 1 - m$. Hence, the equation of the line is $y = mx + 1 - m$.

Now we want the above line to be a tangent to $2x^2 + y^2 = 1$. This means the line should intersect the ellipse at exactly one point. Plug in $y = mx + 1 - m$ and the condition that the line intersects the ellipse at only one point means that the quadratic in $x$ must have only one root which means that the discriminant of the quadratic must be zero.

The quadratic we get is $2x^2 + (mx + 1 - m)^2 = 1$. Rearranging we get $$(m^2+2)x^2 + 2m(1-m)x + m^2 - 2m = 0$$ The discriminant is $4m^2(1-m)^2 - 4(m^2-2m)(m^2+2) = -4m(m-4)$

Setting this to zero, we get the two tangent from $(1,1)$ to the ellipse $2x^2 + y^2 = 1$ are $$y= 1$$ $$y = 4x - 3$$

share|improve this answer
    
Thanks, that's a great answer. Does this technique work for all curves in $R^2$? –  oliland Aug 17 '11 at 15:37
1  
This idea holds good for any conic. This need not be true for arbitrary curves since for instance, the tangent to sin(x) from the point $(1,1)$ intersects the the sine curve at infinite points. –  user17762 Aug 17 '11 at 15:41
    
You're right even if for nice curves (say $C^1$) this is valid locally which means that in a small enough interval... for algebraic curves the contact equation is a polynomial and whenever this polynomial has a root of a multiplicity higher than 1 we have tangency, the other roots are other intersection points. –  marwalix Aug 18 '11 at 6:06

General equation of a line that goes through $P$ is $y-1=m(x-1)$; substituting we get $(m^2+2)x^2-2m(m-1)x+(m-1)^2-1=0$. This quadratic equation in $x$ has a double solution if and only if its discriminant is $0$. This is equivalent to $m(m-4)=0$ (do the computation).

share|improve this answer

Take the derivative of the equation for the ellipse: $$ 4x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0 $$ to get the slope, $\frac{\mathrm{d}y}{\mathrm{d}x}$, of the ellipse at $(x,y)$. The slope of a line going through $(x,y)$ and $(1,1)$ is $\frac{y-1}{x-1}$. Therefore, the points of tangency on the ellipse must satisfy $$ 4x+2y\frac{y-1}{x-1}=0 $$

share|improve this answer
    
One of the points on this tangent is (1,1) - which does not satisfy your equation? –  oliland Aug 17 '11 at 16:02
    
The answer says "points of tangency." The point of tangency of the line $y=1$ is $(0,1)$. –  André Nicolas Aug 17 '11 at 16:21

This is really a question in projective geometry, and has a constructive solution involving only straightedge. To see why it works, you have to know the theory, but here’s how to do it. From your outside point, call it $O$, draw two lines $\ell_1$ and $\ell_2$ each intersecting the ellipse in two points, say $\ell_1$ intersects in $A$ and $A'$, $\ell_2$ intersecting in $B$ and $B'$. Then draw the lines $\overline{AB}$ and $\overline{A'B'}$, intersecting at the point $P_1$ and the lines $\overline{AB'}$ and $\overline{A'B}$ intersecting at the point $P_2$. Then the line $m$ from $P_1$ to $P_2$ intersects your ellipse at the two points of tangency.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.