Ellipse perpendicular distance

In summary: Now we can say that the tangent line is given by x cos(\theta)+ y sin(\theta)= p. We also have that a^2cos^2(\theta)+ b^2sin^2(\theta)= p^2. Now we can solve for cos(\theta) and sin(\theta) in terms of a, b, and p and substitute those into the equation x cos(\theta)+ y sin(\theta)= p to get the tangent line in terms of a, b, and p. We can then use the perpendicular distance formula to find u and v, and show that u^2+ v^2 is a constant.
  • #1

Homework Statement




Prove that the straight line x cos @ + y sin @ = p is a tangent to the ellipse x2/a2 + y2/b2 if a2 cos2@ + b2 sin2@ =p2 .
u and v are the perpendicular distances of a tangent from the two points M(0,ae) and N(0,-ae) respectively. Prove that u2 + v2 is a constant.


2. The attempt at a solution

So far I have: xx1/a2+yy1/b2 -1 = x cos@ + y sin@ - p
I don't know where to go from here and I know in the second part I have to use the perpendicular distance formula but I'm not sure what to sub in and why.
 
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  • #2
Rikendogenz said:

Homework Statement




Prove that the straight line x cos @ + y sin @ = p is a tangent to the ellipse x2/a2 + y2/b2 if a2 cos2@ + b2 sin2@ =p2 .
u and v are the perpendicular distances of a tangent from the two points M(0,ae) and N(0,-ae) respectively. Prove that u2 + v2 is a constant.


2. The attempt at a solution

So far I have: xx1/a2+yy1/b2 -1 = x cos@ + y sin@ - p
I don't know where to go from here and I know in the second part I have to use the perpendicular distance formula but I'm not sure what to sub in and why.
You mean the elliplse [tex]x^2/a^2+ y^2/b^2=1[/tex]. You can simplify that a little by writing it as [tex]b^2x^2+ a^2y^2= a^2b^2[/tex]. Differentiating with respect to x, 2b^2x+ 2a^2yy'= 0 so y'= -b^2x/a^2y. The tangent line at the point [tex](x_1, y_1)[/tex] is given by [tex]y= -(b^2x_1/a^2y_1)(x- x_1)+ y_1[/tex] or b^2x_1x+ a^2y_1y= b^2x_1^2+ a^2y_1^2[/tex].

If we are going to be able to write the coefficients as [tex]cos(\theta)[/itex] and [itex]sin(\theta)[/itex], we must We must have [tex]cos^2(\theta)+ sin^2(\theta)= 1[/tex]. Now we have [tex]b^4x_1^2+ a^4y_1^2[/tex] as the sum of the squares of the coefficients so divide the entire equation by the square root of that:
[tex]\frac{b^2x_1}{\sqrt{b^4x_1^2+ a^4y_1^2}}x+ \frac{a^2y_1}{\sqrt{b^4x_1^2+ a^4y_1^2}}y= \frac{b^2x_1^2+ a^2y_1^2}{\sqrt{b^4x_1^2+ a^4y_1^2}}[/tex]

That is of the form [tex]cos(\theta)x+ sin(\theta)b= p[/tex]
 

1. What is the definition of "Ellipse perpendicular distance"?

The Ellipse perpendicular distance is the shortest distance from any point on an ellipse to its center. It is also referred to as the semi-minor axis of the ellipse.

2. How is the Ellipse perpendicular distance calculated?

The Ellipse perpendicular distance can be calculated using the formula d = b√(1 - (x/a)^2), where d is the perpendicular distance, b is the semi-minor axis, and a is the semi-major axis of the ellipse.

3. What is the significance of the Ellipse perpendicular distance in geometry?

The Ellipse perpendicular distance is important in geometry because it helps determine the shape and size of an ellipse. It is also used in various applications such as satellite orbits, architectural designs, and engineering projects.

4. Can the Ellipse perpendicular distance be negative?

No, the Ellipse perpendicular distance cannot be negative because it is always measured as a positive distance from the center of the ellipse to any point on the ellipse's perimeter.

5. How does the Ellipse perpendicular distance relate to other properties of an ellipse?

The Ellipse perpendicular distance is related to the semi-major axis, semi-minor axis, and eccentricity of an ellipse. It is also used in the calculation of the circumference, area, and focus points of an ellipse.

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