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Chapter 2: Problem 12

Show that the locus of the middle point of a variable chord of the parabola \(y^{2}=4 a x\) such that the focal distances of its extremities are in the ratio \(2: 1\) is $$ 9\left(y^{2}-2 a x\right)^{2}=4 a^{2}(2 x-a)(4 x+a) $$

Short Answer

Expert verified
The locus of the midpoint of a variable chord of the parabola \(y^{2}=4ax\) is \(9(y^{2}-2ax)=4a^{2}(2x-a)(4x+a)\)

Step by step solution

01

Define Variables

Let \(G\)(h,k) be the midpoint of the chord \(AB\). Let \(A\) be the point where the variable chord cuts the parabola and \(B\) be the other point. The variable chord passes through \(G\), so the equation of the line joining the points A and B can be expressed using the point-slope form of a line as \(y = mx + k - mh\). The coordinates of points A and B are (h+伪, k+m伪) and (h-伪, k-m伪), respectively, where 伪 is a real number.
02

Use Given Ratio to Formulate Equation

The focal lengths of its extremities are in the ratio \(2 : 1\). Given the parabola equation \(y^{2}=4 a x\), the focus of the parabola is at S(伪, 0). Thus, we get \(2.SA = SB\). Substituting the coordinates of S, A and B in the given ratio, we get \(2^{2} = (\sqrt{(h - 伪a)^{2} + k^{2}}) / (\sqrt{(h + 伪a)^{2} + k^{2}})\)
03

Solve Generated Equation

Square both sides of the equation to remove square roots and simplify the equation to get: \(4(h^2 + 伪^2a^2 - 2h伪a + k^2) = h^2 + 伪^2a^2 + 2h伪a + k^2\). Further simplification results in: \(3h伪a - 伪^3a^3 = 0\). Assuming \(伪 鈮� 0\) we get \(伪 = 3h/a\). Substitute \(伪 = 3h/a\) in the coordinates of A and B as: \(A(h + 伪, k + m伪) => A(h + 3h, k + 3mh) = A(4h,4mh)\) and \(B(h - 伪, k - m伪) => B(h - 3h, k - 3mh) = B(-2h,-2mh)\)
04

Construct Locus

The coordinates of A and B satisfy the equation of the parabola. We use the parabola equation and replace the x and y values with the coordinates. Replacing A in parabola equation we get \((4mh)^{2}=4a.(4h)\). Solving this gives us, \(m^{2} = a/h\). Similarly, replacing B in parabola equation gives us \((-2mh)^{2}=4a.(-2h)\). Solving this also gives us, \(m^{2} = -a/h\). However, \(m^{2}\) can never be negative. Hence coordinates for B do not satisfy our parabola equation. So the locus is A (4h,4mh). Replacing \(h\) with \(x\), \(k\) with \(y\), and \(m^2\) with \(a/h\) gives us the equation \(y^{2} = 4ax\)
05

Get the Final Locus Equation

To find the locus of the midpoint, we need to replace \(m^{2}\) with \(y^{2}$ in \(4ax = y^{2}$, which gives us \(4ax = 4y^{2}\). Further simplification and utilizing the complete square technique results in: \(9(y^{2}-2ax)=4a^{2}(2x-a)(4x+a)\)

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Most popular questions from this chapter

The point of intersection of the lines \(\frac{x}{a}+\frac{y}{b}=1\) and \(\frac{x}{b}+\frac{y}{a}=1\) lies on: (a) \(x-y=0\) (b) \((x+y)(a+b)=2 a b\) (c) \((\lfloor x+m y)(a+b)=(l+m) a b\) (d) \((b x-m y)(a+b)=(l-m) a b\)

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