/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Let \(F_{1}\) and \(F_{2}\) deno... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(F_{1}\) and \(F_{2}\) denote the foci of the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1(a>b) .\) Suppose that \(P\) is one of the endpoints of the minor axis and angle \(F_{1} P F_{2}\) is a right angle. Compute the eccentricity of the ellipse.

Short Answer

Expert verified
The eccentricity of the ellipse is \(e = \frac{b}{a}\).

Step by step solution

01

Understand the ellipse equation

We begin by noting the standard form of the ellipse equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(a > b\). This implies the ellipse is horizontally oriented, with semi-major axis \(a\) and semi-minor axis \(b\).
02

Identify foci and endpoint of minor axis

The foci \(F_1\) and \(F_2\) of the ellipse are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 - b^2}\). An endpoint of the minor axis is \(P = (0, b)\).
03

Use the right angle condition

The problem states that \(\angle F_1 P F_2 = 90^\circ\). This geometric condition implies that the distance \(|F_1 P|\) and \(|F_2 P|\) must satisfy the Pythagorean theorem, i.e., \(|F_1 P|^2 + |F_2 P|^2 = |F_1 F_2|^2\).
04

Calculate distances

Using the coordinates \(F_1 (-c, 0)\), \(F_2 (c, 0)\), and \(P (0, b)\), we have \(|F_1 P| = \sqrt{(-c)^2 + b^2} = \sqrt{c^2 + b^2}\) and \(|F_2 P| = \sqrt{c^2 + b^2}\). The distance between \(F_1\) and \(F_2\) is \(|F_1 F_2| = 2c\).
05

Solve the equation using Pythagorean theorem

Applying the condition: \((\sqrt{c^2 + b^2})^2 + (\sqrt{c^2 + b^2})^2 = (2c)^2\), simplifies to \(2(c^2 + b^2) = 4c^2\). Solving this equation gives \(c^2 + b^2 = 2c^2\), hence \(b^2 = c^2\).
06

Find eccentricity

Recall the definition of eccentricity for an ellipse is \(e = \frac{c}{a}\). Since \(c^2 = b^2\), \(c = b\). Therefore, the eccentricity \(e = \frac{b}{a}\), where originally \(b < a\). Therefore, \(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{c^2}{a^2}} = \frac{b}{a}\).
07

Conclusion

Since \(c = b\), the eccentricity is \(e = \frac{b}{a}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ellipse Equation
The ellipse equation is fundamental to understanding ellipses. It is expressed in its standard form as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \(a\) is the length of the semi-major axis, and \(b\) is the length of the semi-minor axis. If \(a > b\), the ellipse is elongated horizontally. This equation describes all points \((x, y)\) that make the sum of distances from any point on the ellipse to the two foci constant. In simpler terms, the equation provides the blueprint for an ellipse, ensuring uniformity in its shape. By adjusting the values of \(a\) and \(b\), you can determine the size and orientation of the ellipse.
Foci of an Ellipse
The foci of an ellipse are two special points, denoted as \(F_1\) and \(F_2\), which lie on the major axis. For an ellipse centered at the origin, these points are located at \((\pm c, 0)\), where \(c^2 = a^2 - b^2\). These points are not part of the ellipse itself but are crucial in its definition. The unique feature of an ellipse is that the sum of the distances from any point on the ellipse to these two foci is constant, illustrating a key property defining the shape. This constant relationship is what gives the ellipse its stretched or compressed appearance depending on the values of \(a\) and \(b\). The distance \(c\) is derived using the formula \(c = \sqrt{a^2 - b^2}\), reflecting the fact that the foci move further apart as the ellipse becomes more eccentric.
Minor Axis
The minor axis of an ellipse is the shortest axis of the ellipse. It is perpendicular to the major axis, which is longer. In the standard equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the minor axis has a length of \(2b\). The minor axis, being the shorter axis, is often used to determine the vertical extent of the ellipse when \(a > b\), indicating a horizontally stretched ellipse. The endpoint of the minor axis can be defined and calculated using the coordinates \((0, \pm b)\), given that the center of the ellipse is \((0,0)\). This point is significant in calculations involving angles and distances within the ellipse, which are crucial when applying the Pythagorean theorem to solve related problems.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle of geometry applied in many aspects, including analyzing ellipses. For right-angled triangles, it defines the relationship between the side lengths: the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the context of ellipses, this theorem assists in bridging geometric relationships. For the initial problem, since the angle \(\angle F_1 P F_2\) is a right angle, we use the theorem to affirm that \(|F_1 P|^2 + |F_2 P|^2 = |F_1 F_2|^2\). By substituting known values from the ellipse, such as \(|F_1 P|\), \(|F_2 P|\), and \(|F_1 F_2|\), the relationship facilitates the solution of equations necessary to ascertain other parameters like eccentricity. This effective use of the Pythagorean Theorem underscores its utility beyond triangles, extending its application to more complex geometrical shapes like ellipses.

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