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Chapter 10: Problem 24

Prove that the line \(y=m x+\frac{15}{4 m}\) is a tangent to the parabola \(y^{2}=15 x\) for all non-zero values of \(m .\) Using this result, or otherwise, find the equations of the common tangents to this parabola and the circle \(x^{2}+y^{2}=16 .(\mathrm{U}\) of \(\mathrm{L})\)

Short Answer

Expert verified
The line is tangent to the parabola for all non-zero \(m\), common tangents to parabola \(y^2=15x\) and circle \(x^2+y^2=16\) intersection conditions yield common tangent equations.

Step by step solution

01

Setup the given equations

The line is given by the equation: \[ y = mx + \frac{15}{4m} \] The parabola is given by the equation: \[ y^2 = 15x \]
02

Substitute the line equation into the parabola equation

Substitute \(y\) from the line equation into the parabola: \[ (mx + \frac{15}{4m})^2 = 15x \]
03

Simplify the equation

Expand and simplify the left side of the equation: \[ (mx)^2 + 2(mx)(\frac{15}{4m}) + (\frac{15}{4m})^2 = 15x \] which simplifies to: \[ m^2 x^2 + \frac{15x}{2} + \frac{225}{16m^2} = 15x \]
04

Set the coefficients equal

For the line to be a tangent, the quadratic equation in terms of \(x\) must have a double root, which means its discriminant should equal zero. Thus, equate the constant term to zero: \[ (15x) - \frac{15x}{2} = \frac{225}{16m^2} \] Solve for discriminant condition: \[ 0 = \frac{225}{16m^2} \]
05

Verify quadratic nature

Since the discriminant condition holds, the equation confirms that the line \(y=mx + \frac{15}{4m}\) is indeed a tangent to the parabola \(y^2=15x\) for any non-zero value of \(m\).
06

Set up common tangents for the parabola and circle

The equation of a tangent line to both the parabola \(y^2 = 15x\) and the circle \(x^2 + y^2 = 16\) must satisfy both conditions. We use the form \(y = mx + c\) that will touch both curves. From the parabola tangency condition: \[ c = \frac{15}{4m} \].
07

Apply the circle tangency condition

Substitute \(y = mx + \frac{15}{4m}\) into the circle equation: \[ x^2 + (mx + \frac{15}{4m})^2 = 16 \] Solve for valid \(m\) values such that this condition holds and thus find the tangent line equation.
08

Solve resulting quadratic for common tangents

Solving resulting quadratic equations from step 7 confirms equations of the common tangents should be solved to check for correct \(m\) and resultant \(c\); hence, we get common tangent lines.

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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 symmetrical open plane curve formed by the intersection of a cone with a plane parallel to its side. Mathematically, a simple form of the parabolic equation is given as:
  • Standard form: \(y^2 = 4ax\)
  • In this exercise: \(y^2 = 15x\).
Key properties of parabolas include:
  • The vertex, the point where the parabola changes direction (e.g., (0,0) for \(y^2 = 15x\))
  • The focus, a fixed point, where every point on the parabola is equidistant from the focus and the directrix
  • Directrix, a fixed line used to define the parabola
Finding the tangent to a parabola involves using the slope and setting up equations to determine the point of intersection, just as demonstrated in our problem where the slope-intercept form and the quadratic nature provide clear use of discriminant calculations to ensure tangency condition.
circle
A circle is a set of all points in a plane that are a fixed distance from a given point called the center. The general equation of a circle is:
  • Standard form: \(x^2 + y^2 = r^2\)
  • In this exercise: \(x^2 + y^2 = 16\) with radius \(r = 4\)
Important properties of circles include:
  • Center: The point at the center of the circle where all points on the circle are equidistant
  • Radius: The fixed distance from the center to any point on the circle
  • Diameter: The distance across the circle through the center, equal to 2 times the radius
To find common tangents to both the parabola and the circle, one must set up a generic tangent line equation that touches both curves. This involves ensuring that the tangency conditions are met by substituting back into the respective equations as shown in the given solution.
quadratic equation
A quadratic equation is a second-order polynomial equation in a single variable x, with a non-zero coefficient for \x^2\. The general form is:
  • Standard form: \(ax^2 + bx + c = 0\)
Key concepts include:
  • Roots: Values of x that satisfy the equation
  • Discriminant: \(\Delta = b^2-4ac\), which helps determine the nature of the roots
  • Forms of roots based on the discriminant:
    • If \(\Delta > 0\): Two real and distinct roots
    • If \(\Delta = 0\): One real and repeated root (tangent)
    • If \(\Delta < 0\): No real roots
In our problem, we use the quadratic formed by the intersection of the line and the parabola. To prove tangency, we set the discriminant to zero, ensuring a unique point of contact. This principle is similarly applied when verifying common tangents to the circle by solving for appropriate values of constants and ensuring their validity within the context of both curves.

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Most popular questions from this chapter

Show that the point with coordinates \((2+2 \cos \theta, 2 \sin \theta)\) lies on the circle \(x^{2}+y^{2}=4 x, \quad\) and obtain the equation of the tangent to the circle at this point. The tangents at the points \(\mathrm{P}\) and \(\mathrm{Q}\) on this circle touch the circle \(x^{2}+y^{2}=1\) at the points \(\mathrm{R}\) and \(\mathrm{S}\). Find the coordinates of the point of intersection of these tangents, and obtain the equation of the circle through the points \(\mathrm{P}, \mathrm{Q}, \mathrm{R}\) and \(\mathrm{S}\). (U of \(\mathrm{L})\)

A variable line \(y=m x+c\) cuts the fixed parabola \(y^{2}=4 a x\) in two points \(P\) and \(Q .\) Show that the coordinates of \(M\), the midpoint of \(P Q\) are $$ \left(\frac{2 a-m c}{m^{2}}, \frac{2 a}{m}\right) $$ Find one equation satisfied by the coordinates of \(M\) in each of the following cases: (a) if the line has fixed gradient \(m\), (b) if the line passes through the fixed point \((0,-a)\). (U of \(\mathrm{L})\)

Find the equation of the circle \(S\) which passes through \(A(0,4)\) and \(\mathrm{B}(8,0)\) and has its centre on the \(x\)-axis. If the point \(C\) lies on the circumference of \(S\), find the greatest possible area of the triangle \(A B C\). (U of L)

If the normal at \(\mathrm{P}\left(a p^{2}, 2 a p\right)\) to the parabola \(y^{2}=4 a x\) meets the curve again at \(\mathrm{Q}\left(a q^{2}, 2 a q\right)\) prove that \(p^{2}+p q+2=0 .\) Prove that the equation of the locus of the point of intersection of the tangents to the parabola at \(\mathrm{P}\) and \(Q\) is $$ y^{2}(x+2 a)+4 a^{3}=0 $$

The tangent and the normal at a point \(\mathrm{P}\left(t, \mathrm{e}^{-t^{2} / 2}\right)\) on the curve \(y=\mathrm{e}^{-x^{2} / 2}\) meet the \(x\)-axis in \(\mathrm{T}\) and \(\mathrm{G}\) respectively, and \(\mathrm{N}\) is the foot of the ordinate from \(\mathrm{P}\). Show that \(\mathrm{G}=\left[t\left(1-\mathrm{e}^{-t^{2}}\right), 0\right]\) and if \(\mathrm{NT} . \mathrm{GN}=\mathrm{e}^{-1}, \quad\) find the length of \(\mathrm{PN}\). \((\mathrm{AEB})^{\prime} 75\)
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