conics help (1 Viewer)

[VJ30]

can anyone help me with this question please?

find the equations of two tangents to the ellipse 16x^2 + 25y^2= 400 which are parallel to the line y=x+2 thanks

i found the condition that tan(theta)= -4/5 i cant go further some pleease show me the steps..

 

[Carrotsticks]

You need to find the general equation of all straight lines that are tangential to the ellipse. To do this:

1. Find the MOST GENERAL equation of a straight line, in this case y=mx+b.

2. Sub into the ellipse to acquire a quadratic in terms of x.

3. Let the discriminant be equal to zero. Deduce a condition.

4. Sub this condition back into the equation of the line.

5. Now we want the tangent(s) to be parallel to y=x+2 ie: gradient of 1, so we let m=1.

6. Upon subbing m=1, you should get two equations.

 

[Aesytic]

if tan theta = -4/5, then you know that theta is either pi - arctan(4/5) or 2pi - arctan(4/5) since tan is negative in the second and fourth quadrants
from this you can work out the points where they touch the ellipse with the parametric equations x=5cos(theta) and y=4sin(theta), then use the point gradient formula with gradient=1 to find the 2 tangents to the ellipse

edit: assuming that by condition you mean tan(theta) must equal -4/5 for the tangent to the ellipse to have a gradient of 1, and that by theta you mean theta in the parametric equations of the ellipse

 

[Fus Ro Dah]

if tan theta = -4/5, then you know that theta is either pi - arctan(4/5) or 2pi - arctan(4/5) since tan is negative in the second and fourth quadrants
from this you can work out the points where they touch the ellipse with the parametric equations x=5cos(theta) and y=4sin(theta), then use the point gradient formula with gradient=1 to find the 2 tangents to the ellipse

edit: assuming that by condition you mean tan(theta) must equal -4/5 for the tangent to the ellipse to have a gradient of 1, and that by theta you mean theta in the parametric equations of the ellipse

Your method works, but it doesn't exactly give a general equation. Suppose you wanted to calculate a similar condition but with gradient 2, then you would have to do that all over again whereas Carrotsticks' method has a general form where you simply substitute m=k for any real k.

 

[RealiseNothing]

I differentiated the equation of an ellipse, then let it equal 1. Re-arranged this to make 'x' the subject and substituted this into the ellipse to find 'x' and 'y' values. Then just used point-gradient formula to find the equation of the two tangents.

 

[Aesytic]

Your method works, but it doesn't exactly give a general equation. Suppose you wanted to calculate a similar condition but with gradient 2, then you would have to do that all over again whereas Carrotsticks' method has a general form where you simply substitute m=k for any real k.

yeah, i felt that that would be the quickest way to answer OP's question. if there were more parts to it, like finding tangents with gradient 2, then i would've done it the same way as carrotsticks