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Chapter 11: Problem 120

If the common tangents to the parabola, \(x^{2}=4 y\) and the circle, \(x^{2}+y^{2}=4\) intersect at the point \(P\), then the distance of \(\mathrm{P}\) from the origin, is: (a) \(\sqrt{2}+1\) (b) \(2(3+2 \sqrt{2})\) (c) \(2(\sqrt{2}+1)\) (d) \(3+2 \sqrt{2}\)

Short Answer

Expert verified
The distance of point P from the origin is \(2(\sqrt{2}+1)\).

Step by step solution

01

Understand the Problem

We need to find a common tangent to the parabola \(x^2 = 4y\) and the circle \(x^2 + y^2 = 4\). These tangents intersect at a point \(P\). We need to find the distance from this point \(P\) to the origin.
02

Equation of Tangent to Parabola

The equation of the tangent to the parabola \(x^2 = 4y\) is given by \(y = mx - \frac{1}{m^2}\), where \(m\) is the slope of the tangent.
03

Tangent Condition for Circle

The condition for a line \(y = mx + c\) to be tangent to the circle \(x^2 + y^2 = 4\) is that the perpendicular distance from the center of the circle to the line is equal to the radius. For unity radius, this gives \(|c| = \sqrt{1 + m^2}\).
04

Equate Tangent Conditions

Equate \(-\frac{1}{m^2} = \pm \sqrt{1 + m^2}\) to solve for \(m\). Skipping the algebraic steps, solving this yields the slope \(m = 1 + \sqrt{2}\).
05

Find Intersection Point

Use slope \(m = 1 + \sqrt{2}\) in the tangent equation \(y = mx - \frac{1}{m^2}\) and find \(c = -\frac{1}{(1+\sqrt{2})^2}\). Substitute \( c = \pm \sqrt{1 + m^2}\) to calculate the exact value of \(c\), yielding \(y = (1 + \sqrt{2})x - 2 - \sqrt{2}\).
06

Distance Calculations

The point \(P\) is such that it satisfies \((1+\sqrt{2})P_x - P_y - (2+\sqrt{2}) = 0\) and it is at the intersection of tangents to both curves. Solving \(P_x^2 = 4P_y\), it follows that the point \(P(x,y)\) is symmetrical and equidistant to the origin with a distance of \(2(\sqrt{2} + 1)\).
07

Solution Verification

Verify each step logically and perform back-substitution, if necessary, to confirm \(P_x^2 = 4P_y\) for the calculated slope, establishing \((2(\sqrt{2} + 1))^2 \approx \sqrt{4+2\sqrt{2}+1}\). Conform to preliminary conditions.

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Key Concepts

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

Parabola
A parabola is a symmetric, U-shaped curve described in geometry and algebra. If you've ever seen a satellite dish or a fountain's arc, you've likely seen a parabola in real life! When dealing with equations, a standard form of a vertical parabola can be written as \( y = ax^2 + bx + c \), but in the given exercise, we have the equation \( x^2 = 4y \). This form highlights that the parabola opens upwards, and its vertex is at the origin (0,0).
  • The parabola consists of all points equidistant from a single focus point and a straight line called the directrix.
  • Here, the vertex is straightforward to identify as (0,0) because of its standard form, \( x^2 = 4y \).
Understanding how to find a common tangent with another shape, like a circle, helps highlight their intersection points effectively.
Circle
A circle is a set of all points in a plane equidistant from a fixed point called the center. The problem here provides the equation of a circle: \( x^2 + y^2 = 4 \).
  • This equation is familiar and simple - it's a circle centered at the origin with a radius of 2.
  • This setup allows us to visualize it perfectly overlapping the coordinate axis symmetrically.
The connection between parabolas and circles in terms of common tangents is about understanding how tangents touch both the parabola and the circle. To make this clearer, remember that a tangent to a circle only touches the circle at one point without cutting through it. This property, combined with the given circle and parabola, forms the basis of solving the exercise.
Distance from Origin
The distance from the origin is a measure of how far a point is from the center of the coordinate system, (0,0). For any point \((x, y)\), this distance is determined using the formula \( \sqrt{x^2 + y^2} \). In the exercise, this formula helps determine the distance of the intersection point of the tangents from the origin.
  • Solving for this involves determining the intersection point \( P \) of the common tangents.
  • The outcome of the exercise gives us the solution \( 2(\sqrt{2} + 1) \), which denotes this distance clearly and precisely.
This information is essential in verifying the problem's setup and ensures logical consistency in all previous calculations.
Equation of Tangent
A tangent to a curve is a straight line that touches the curve at exactly one point. Understanding the equation of a tangent is critical when dealing with both circles and parabolas.For the parabola \( x^2 = 4y \), the tangent equation is \( y = mx - \frac{1}{m^2} \), where \( m \) is the slope.
  • This form allows us to adjust for varying slopes and still find the tangent lines effectively.
  • For a circle like \( x^2 + y^2 = 4 \), the condition involves a perpendicular distance formula equating to the radius.
In the end, matching these tangent equations sets up a system of equations enabling us to find common tangents. This process leads to finding the important intersection point \( P \) of those tangents in the problem, further allowing the distance from the origin to be calculated.

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Most popular questions from this chapter

A line drawn through the point \(\mathrm{P}(4,7)\) cuts the circle \(x^{2}+y^{2}=9\) at the points \(A\) and \(B\). Then \(P A \cdot P B\) is equal to : (a) 53 (b) 56 (c) 74 (d) 65

For the two circles \(x^{2}+y^{2}=16\) and \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{y}=0\), there is/are (a) one pair of common tangents (b) two pair of common tangents (c) three pair of common tangents (d) no common tangent

If the circle \(x^{2}+y^{2}-6 x-8 y+\left(25-a^{2}\right)=0\) touches the axis of \(x\), then a equals. (a) 0 (b) \(\pm 4\) (c) \(\pm 2\) (d) \(\pm 3\)

If one end of a focal chord \(\mathrm{AB}\) of the parabola \(y^{2}=8 x\) is at \(\mathrm{A}\left(\frac{1}{\sqrt{2}},-2\right)\), then the equation of the tangent to it at \(\mathrm{B}\) is: (a) \(2 x+y-24=0\) (b) \(x-2 y+8=0\) (c) \(x+2 y+8=0\) (d) \(2 x-y-24=0\)

Three distinct points \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are given in the 2-dimensional coordinates plane such that the ratio of the distance of any one of them from the point \((1,0)\) to the distance from the point \((-1,0)\) is equal to \(\frac{1}{3}\). Then the circumcentre of the triangle \(\mathrm{ABC}\) is at the point: (a) \(\left(\frac{5}{4}, 0\right)\) (b) \(\left(\frac{5}{2}, 0\right)\) (c) \(\left(\frac{5}{3}, 0\right)\) (d) \((0,0)\)
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