Angle Between Tangents Drawn to Ellipse 3x2+5y2=15 from (2,2)

[Saitama]

Homework Statement

The angle between the tangents drawn from the point (2,2) to the ellipse, 3x²+5y²=15 is:
a)##\pi##/6
b)##\pi##/4
c)##\pi##/3
d)##\pi##/2

Homework Equations

The Attempt at a Solution

To find the equation of tangents, I need to use the following formula:
[tex]\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\right)=\left(\frac{xx_1}{a^2}+\frac{yy_1}{b^2}-1\right)^2[/tex]
(x1,y1 is point from which the tangents are drawn to the ellipse.)
This will give me an equation of 2 degree and using this, separate equations of tangents can be found. To find the angle between the two tangents, I will use the following formula.
[tex]\tan\theta=\frac{m_1-m_2}{1+m_1m_2}[/tex]
(m1 and m2 are the slopes of the two tangents)
But going through this process is too much work, is their any simpler method? Is their any trick to do the above question?

Any help is appreciated. Thanks!

 

[yy205001]

The angle between the tangent at (2,2) from the ellipse and x-axis??

 

[Saitama]

The angle between the tangent at (2,2) from the ellipse and x-axis??

Please re-read the question. It asks the angle between the two tangents drawn from (2,2) to the ellipse.

 

[HallsofIvy]

I hate memorizing formulas like the one you give. Instead, note that, from the equation of the ellipse, 6x+ 10yy'= 0 so that y'= -(5/3)(x/y) at each (x, y) on the ellipse. In particular, if we take [itex](x_0, y_0)[/itex] to be the point on the ellipse at which a line through (2, 2) is tangent to the ellipse, then that line must have equation [itex]y= -(5/3)(x_0/y_0)(x- 2)+ 2[/itex]. Since [itex](x_0, y_0)[/itex] is itself on that line, we must have [itex]y_0= -(5/3)(x_0/y_0)(x- 2)+ 2[/itex] or, multiplying by [itex]y_0[/itex], [itex]y_0^2= -(5/3)x_0(x_0- 2)+ 2y_0= -(5/3)x_0^2+ (10/3)x_0+ 2y_0[/itex]. We can write that as [itex](5/3)x_0^2+ (10/3)x_0+y_0^2- 2y_0= 0[/itex].

Of course, [itex](x_0, y_0)[/itex] also lies on the ellipse so we also have [itex]3x_0^2+ 5y_0^2= 15[/itex]. That gives two quadratic equations to solve giving two points at which the two lines tangent to the ellipse through (2, 2) touch the ellipse. Now that you know two points on each tangent line, you can find their slope, [itex]\tan(\theta)[/itex], where [itex]\theta[/itex] is the angle the line makes with the x-axis.

Finally, to find the angle between them use the identity
[tex]tan(\theta_1- \theta_2)= \frac{tan(\theta_1)- tan(\theta_2)}{1+ tan(\theta_1)tan(\theta_2)}[/tex]

 

[Saitama]

I hate memorizing formulas like the one you give. Instead, note that, from the equation of the ellipse, 6x+ 10yy'= 0 so that y'= -(5/3)(x/y) at each (x, y) on the ellipse. In particular, if we take [itex](x_0, y_0)[/itex] to be the point on the ellipse at which a line through (2, 2) is tangent to the ellipse, then that line must have equation [itex]y= -(5/3)(x_0/y_0)(x- 2)+ 2[/itex]. Since [itex](x_0, y_0)[/itex] is itself on that line, we must have [itex]y_0= -(5/3)(x_0/y_0)(x- 2)+ 2[/itex] or, multiplying by [itex]y_0[/itex], [itex]y_0^2= -(5/3)x_0(x_0- 2)+ 2y_0= -(5/3)x_0^2+ (10/3)x_0+ 2y_0[/itex]. We can write that as [itex](5/3)x_0^2+ (10/3)x_0+y_0^2- 2y_0= 0[/itex].

Of course, [itex](x_0, y_0)[/itex] also lies on the ellipse so we also have [itex]3x_0^2+ 5y_0^2= 15[/itex]. That gives two quadratic equations to solve giving two points at which the two lines tangent to the ellipse through (2, 2) touch the ellipse. Now that you know two points on each tangent line, you can find their slope, [itex]\tan(\theta)[/itex], where [itex]\theta[/itex] is the angle the line makes with the x-axis.

Finally, to find the angle between them use the identity
[tex]tan(\theta_1- \theta_2)= \frac{tan(\theta_1)- tan(\theta_2)}{1+ tan(\theta_1)tan(\theta_2)}[/tex]

Thanks for the help, HallsofIvy!

But this still is a long process. I would like to know if there is any trick to do the above question by looking through the options given.

 

[mfb]

A careful sketch could work. 2 is close to the largest x- (~2.2) and the largest y-value (~1.7). If that would be exact, the angle would be ##\frac{\pi}{2}##. It is not, but the deviation is not so large, so I would prefer ##\frac{\pi}{2}## over ##\frac{\pi}{3}##

 

[utkarshakash]

For solving these kinds of questions where you have to find the angle between two tangents of an ellipse (especially the MCQ), you should first check the possibility of ∏/2. Why? Because there is a very simple method to do so. OK here's it.

You must know that the point of intersection of pair of perpendicular tangents to the ellipse lies on the director circle the equation of whose is given by [itex]x^2+y^2=a^2+b^2[/itex] where a and b have usual meanings. You have the point(2,2) with yourself. You also know a and b. So why don't you go ahead and check whether that point lies on director circle. If it lies on it then you already know the answer.

 

[Saitama]

For solving these kinds of questions where you have to find the angle between two tangents of an ellipse (especially the MCQ), you should first check the possibility of ∏/2. Why? Because there is a very simple method to do so. OK here's it.

You must know that the point of intersection of pair of perpendicular tangents to the ellipse lies on the director circle the equation of whose is given by [itex]x^2+y^2=a^2+b^2[/itex] where a and b have usual meanings. You have the point(2,2) with yourself. You also know a and b. So why don't you go ahead and check whether that point lies on director circle. If it lies on it then you already know the answer.

Great! Thanks, that was really helpful.