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Chapter 11: Problem 180

If \(P_{1}\) and \(P_{2}\) are two points on the ellipse \(\frac{x^{2}}{4}+y^{2}=1\) at which the tangents are parallel to the chord joining the points \((0,1)\) and \((2,0)\), then the distance between \(P_{1}\) and \(P_{2}\) is [Online May 12, 2012] (a) \(2 \sqrt{2}\) (b) \(\sqrt{5}\) (c) \(2 \sqrt{3}\) (d) \(\sqrt{10}\)

Short Answer

Expert verified
The distance between points is \(\sqrt{10}\).

Step by step solution

01

Find the Slope of the Chord

The points given are \((0,1)\) and \((2,0)\). The slope \(m\) of the line joining these two points is calculated by using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Thus, the slope is \(m = \frac{0 - 1}{2 - 0} = -\frac{1}{2}\).
02

Tangent Slope Formula for Ellipse

For the ellipse \(\frac{x^2}{4} + y^2 = 1\), the slope of the tangent at any point \((x,y)\) is given by the derivative \(\frac{dy}{dx} = -\frac{x}{2y}\). The condition that the tangents are parallel means \(-\frac{x}{2y} = -\frac{1}{2}\).
03

Solve for Points on the Ellipse

From the tangent slope formula \(-\frac{x}{2y} = -\frac{1}{2}\), we get \(x = y\). Substitute \(x = y\) into the ellipse equation. \[ \frac{y^2}{4} + y^2 = 1 \Rightarrow \frac{5y^2}{4} = 1 \Rightarrow y^2 = \frac{4}{5} \Rightarrow y = \pm \frac{2}{\sqrt{5}} \].
04

Calculate Corresponding x-values

Since \(x = y\), the corresponding \(x\) values are also \(x = \pm \frac{2}{\sqrt{5}}\). So the points are \(\left(\frac{2}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\) and \(\left(-\frac{2}{\sqrt{5}}, -\frac{2}{\sqrt{5}}\right)\).
05

Find the Distance Between Points

The distance \(d\) between the points \(\left(\frac{2}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\) and \(\left(-\frac{2}{\sqrt{5}}, -\frac{2}{\sqrt{5}}\right)\) is computed using the distance formula \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). Substituting the values, we have: \[ d = \sqrt{\left(-\frac{2}{\sqrt{5}} - \frac{2}{\sqrt{5}}\right)^2 + \left(-\frac{2}{\sqrt{5}} - \frac{2}{\sqrt{5}}\right)^2} = \sqrt{\left(-\frac{4}{\sqrt{5}}\right)^2 + \left(-\frac{4}{\sqrt{5}}\right)^2} \].
06

Simplify the Expression

Simplifying further: \[ d = \sqrt{2 \left( \frac{16}{5} \right)} = \sqrt{\frac{32}{5}} = \frac{4\sqrt{2}}{\sqrt{5}} = \sqrt{10} \]. Thus, the distance is \(\sqrt{10}\).

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Key Concepts

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

Tangent to Ellipse
When we talk about the tangent to an ellipse, we refer to a line that touches the ellipse at exactly one point without crossing it. At this point, the tangent line is perpendicular to the radius drawn to the point of tangency on the ellipse. Mathematically, the slope of this tangent line to the ellipse \( \frac{x^2}{4} + y^2 = 1 \) can be found using calculus techniques. The formula for the slope of the tangent at any point \((x, y)\) on the ellipse is given by finding the derivative \( \frac{dy}{dx} = -\frac{x}{2y} \). This derivative comes from implicit differentiation of the ellipse equation. Understanding the slope formula is crucial because, in problems like the one provided, we may need to find points on the ellipse where the tangents have specific slopes, such as being parallel to a given line. Knowing when two lines are parallel鈥攊.e., they have the same slope鈥攈elps solve many geometry problems involving ellipses.
Distance Formula
The distance formula is a mathematical tool used to calculate the distance between two points in a plane. This formula is derived from the Pythagorean theorem and is an essential component of analytical geometry. Given two points, \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between them is given by:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]In this problem, after finding points \(P_1\) and \(P_2\) on the ellipse, we apply the distance formula to determine how far apart these points are. Recognizing when to use this formula is vital for solving problems involving distances on a coordinate plane. In the context of ellipses, it can provide solutions for finding lengths of chords or verifying specific conditions about point positions on the ellipse.
Slope of a Line
The slope of a line is a measure of its steepness and direction. Calculating the slope is an essential part of understanding lines in coordinate geometry. The slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is determined using:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]This concept is applicable when determining parallel lines. In this exercise, we find that the slope of the chord (a line between two points) joining the points \((0, 1)\) and \((2, 0)\) is \(-\frac{1}{2}\). This slope is crucial because the tangents to the ellipse must have the same slope to be parallel to the chord. Recognizing relationships between slopes allows for solving geometric problems efficiently, particularly when dealing with shapes like ellipses.
Chord of an Ellipse
A chord of an ellipse is a line segment whose endpoints both lie on the ellipse. Understanding chords is important for solving problems related to geometry and calculus involving ellipses. In this specific exercise, identifying and constructing a chord that matches certain conditions, such as having parallel tangents, is a crucial step.To solve the given problem, we needed to determine where the tangents to the ellipse are parallel to the chord. This involved equating the slope of the chord, found to be \(-\frac{1}{2}\), to the slope of the tangent line to the ellipse. By substituting and solving, we found the points on the ellipse (where the tangents are parallel to the chord), and calculated the straight-line distance between them using the distance formula. Chords help in the visualization and solving of geometric problems related to ellipses, enabling the application of algebraic methods to geometric figures.

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Most popular questions from this chapter

The set of all real values of \(\lambda\) for which exactly two common tangents can be drawn to the circles \(x^{2}+y^{2}-4 x-4 y+6=0\) and \(x^{2}+y^{2}-10 x-10 y+\lambda=0\) is the interval: (a) \((12,32)\) (b) \((18,42)\) (c) \((12,24)\) (d) \((18,48)\)

A variable circle passes through the fixed point \(A(p, q)\) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is (a) \((y-q)^{2}=4 p x\) (b) \((x-q)^{2}=4 p y\) (c) \((y-p)^{2}=4 q x\) (d) \((x-p)^{2}=4 q y\)

The area (in sq. units) of an equilateral triangle inscribed in the parabola \(y^{2}=8 x\), with one of its vertices on the vertex of this parabola, is: (a) \(64 \sqrt{3}\) (b) \(256 \sqrt{3}\) (c) \(192 \sqrt{3}\) (d) \(128 \sqrt{3}\)

Let \(C\) be the circle with centre at \((1,1)\) and radius \(=1\). If \(T\) is the circle centred at \((0, y)\), passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to (a) \(\frac{1}{2}\) (b) \(\frac{1}{4}\) (c) \(\frac{\sqrt{3}}{\sqrt{2}}\) (d) \(\frac{\sqrt{3}}{2}\)

Let \(L_{1}\) be a tangent to the parabola \(y^{2}=4(x+1)\) and \(L_{2}\) be a tangent to the parabola \(y^{2}=8(x+2)\) such that \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\) intersect at right angles. Then \(L_{1}\) and \(L_{2}\) meet on the straight line : (a) \(x+3=0\) (b) \(2 x+1=0\) (c) \(x+2=0\) (d) \(x+2 y=0\)
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