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Chapter 2: Problem 4

Two equal parabolas have the same vertex and their axes are at right angle. Prove that their common tangent touches each at the end of a latus rectum.

Short Answer

Expert verified
The proof for the problem involves a series of steps including understanding the concept of a parabola, latus rectum, and tangent line, defining our two parabolas, finding their common tangent lines, and finally proving that these tangent lines intersect with the end points of the latus rectum. In the end, we find that the lines \(y = x + a\) and \(y = -x + a\) are the common tangents that fulfill the given conditions.

Step by step solution

01

Understand the problem

Firstly, you need to understand that a latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, passing through the focus and with its endpoints on the parabola. A common tangent means the same tangent line for both parabolas.
02

Define the parabolas

Let's define the parabolas with the following standard forms; for parabola 1: \(y^2 = 4ax\) and for parabola 2: \(x^2 = 4ay\), where a is the distance from the vertex to the focus (also half the length of the latus rectum) and x and y are their respective axes.
03

Find the common tangent

For a line to be tangent to both parabolas, it needs to satisfy the tangent condition for both equations. The line \(y = mx + c\) is a tangent to the parabola \(x^2 = 4a^2y\) if \(c = a/m\). Similarly, the line is a tangent to the parabola \(y^2 = 4a^2x\) if \(c^2 = 4*a^2(1+m^2)\). By equating both equations for \(c\), we get that \(m^2 = 1\), so \(m = 1\) or \(m = -1\). Therefore, the equations of the lines tangent to both parabolas are \(y = x + a\) and \(y = -x + a\).
04

Prove the relation with the latus rectum

Now, the latus rectum of the parabolas end at the points where \(x = a\) and \(y = a\) respectively. Substituting these values into the equations of the tangent lines, if the tangents go through these points, it means the lines touch each parabola at the end of a latus rectum. The proof is complete if both tangent lines fulfill this condition.

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Most popular questions from this chapter

Line \(L\) has intercepts \(a\) and \(b\) on the co-ordinate axes, when the axes are rotated through a given angle; keeping the origin fixed, the same line has intercepts \(p\) and \(q\), then : (a) \(a^{2}+b^{2}=p^{2}+q^{2}\) (b) \(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{p^{2}}+\frac{1}{q^{2}}\) (c) \(a^{2}+p^{2}=b^{2}+q^{2}\) (d) \(\frac{1}{a^{2}}+\frac{1}{p^{2}}=\frac{1}{b^{2}}+\frac{1}{q^{2}}\)

Let a given line \(L_{1}\) intersect the \(x\) and \(y\)-axes at \(P\) and \(Q\) respectively. Let another line \(L_{2}\) perpendicular to \(L_{1}\), cut the \(x\) and \(y\)-axes at \(R\) and \(S\), respectively. Show that the locus of the point of intersection of the lines \(P S\) and \(Q R\) is a circle passing through the origin. (iIT 87,3 )

The straight line \(y=x-2\) rotates about a point where it cuts \(x\)-axis and becomes perpendicular on the straight line \(a x+b y+c=0\) then its equation is : (a) \(a x+b y+2 a=0\) (b) \(a y-b x+2 b=0\) (c) \(a x+b y+2 b=0\) (d) none of these

Consider the equation \(y-y_{1}=m\left(x-x_{1}\right)\). If \(m\) and \(x_{1}\) are fixed and different lines are drawn for different values of \(y_{1}\), then (a) the lines will pass through a fixed point (b) there will be a set of parallel lines (c) all the lines intersect the line \(x=x_{1}\) (d) all the lines will be parallel to the line \(y=x_{1}\)

A variable chord \(P Q\) of the parabola \(y^{2}=4 x\) is drawn parallel to the line \(y=x\). If the parameters of the points \(P\) and \(Q\) on the parabola are \(p\) and \(q\) respectively show that \(p+q=2\) Also show that the locus of the point of intersection of the normals at \(P\) and \(Q\) is \(2 x-y=12\)
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