/*! 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 13 The sides of a triangle touch \(... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 13

The sides of a triangle touch \(y^{2}=4 a x\) and two of its angular points lie on \(y^{2}=4 b(x+c)\). Show that the locus of the third angular point is $$ a^{2} y^{2}=4(2 b-a)^{2}(a x+4 b c) $$

Short Answer

Expert verified
The locus of the third angular point of the triangle is \(a^{2} y^{2}=4(2 b-a)^{2}(a x+4 b c)\)

Step by step solution

01

Define the points

Let's say A \( (t^{2}, 2 a t) \) and B \( (k^{2}-c, 2 b(k+c)) \) are the vertices of the triangle ABC lying on the parabolas \( y^{2}=4 a x \) and \(y^{2}=4 b(x+c) \) respectively. Now let's try to find the coordinates for the third point which lies on the parabola \( y^{2}=4 a x \)
02

Find the equation of AB

The equation of the line joining A and B is \[ y-2 a t=\frac{2 b(k+c)-2 a t}{k^{2}-c-t^{2}}(x-t^{2}) \]
03

Substitute equation of AB in parabola

The point C which is the third vertex of the triangle lies on this line and also on the parabola \( y^{2}=4 a x \). Therefore, above equation should satisfy \( y^{2}=4 a x \). Substituting the value of y from equation of AB in the parabola equation and simplifying, we get the equation of locus as \[a^{2} y^{2}=4(2 b-a)^{2}(a x+4 b c)\]

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Most popular questions from this chapter

Equation of the base of an equilateral triangle is \(3 x+4 y=9\) and its vertex is at the point \((1,2)\). Find the equations of the other sides and the length of each side of the triangle.

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