# Parabolas Set 3

**Example – 5**

From a variable point *P* on a fixed normal to the parabola \({y^2} = 4ax,\) two more normals are drawn to the parabola to intersect it at *Q* and *R*. Show that the variable chord *QR* will have a fixed slope.

**Solution:** Let *P* be thepoint (*h, k*). An arbitrary normal to the given parabola can be written as \[y = mx - 2am - a{m^3}\] and if this passes through *P*, we have

\[k = mh - 2am - a{m^3} \qquad \dots \left( 1 \right)\]

Now,this cubic has three roots, say \({m_1},\,\,{m_2}\;{\rm{ and}}\;{\rm{ }}{m_3},\) and of which one say \({m_1}\), is fixed, equal to the slope of the fixed normal.The other two slopes are variable and thus the points of intersection of the other two normals through \(P\,\,\{Q(am_2^2,\, - 2a{m_2}){\rm{\;\; }}and{\rm{ \;\;}}R(am_3^2,\, - 2a{m_3})\} \) are also variable.

From (1), we have

\[{m_1} + {m_2} + {m_3} = 0 \qquad \dots \left( 2 \right)\]

The slope of *QR* is

\[\begin{align}m&= \frac{{2a({m_2}- {m_3})}}{{a(m_3^2 - m_2^2)}}\\\\&= \frac{2}{{ -({m_2} + {m_3})}}\\\\&=\frac{2}{{{m_1}}}\qquad \qquad \qquad\left( {{\rm{from }}\left( 2 \right)} \right)\end{align}\]

This shows that m, the slope of QR, is fixed since \({m_1}\) is fixed.

**Example – 6**

Find the locus of the point of intersection of those normals to the parabola \({x^2} = 8y\) which are at right angles to each other.

**Solution:** We have already discussed a very similar question earlier. The point of including this question here is that the equation to the parabola has been given not in the standard form \({y^2} = 4ax\) but in the form \({x^2} = 4ay:\) this means that in all the formulae that we use, *x* and *y* must be interchanged.

Thus,the equation of an arbitrary normal to \({x^2}= 8y\) can be written as

\[x = my - 4m - 2{m^3}\]

If this passes through \(P(h,k),\) we obtain

\[h = km - 4m - 2{m^3} \qquad \quad \dots \left( 1 \right)\]

This cubic has three roots, say \({m_1},\,{m_2}{\rm{ }}\;and\;{\rm{ }}{m_3}.\) Since two of the normals are perpendicular, we can take \({m_1}{m_2}\) equal to –1.

From (1), we have

\[\begin{align}& {m_1} + {m_2} +{m_3} = 0\\\\&{m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1} = - \frac{k}{2} + 2\\\\&{m_1}{m_2}{m_3}= - \frac{h}{2}\end{align}\]

Using \({m_1}{m_2} = - 1\) and these three equations, arelation involving only *h* and *k* can easily be obtained:

\[{h^2} - 2k + 12 = 0\]

Thus,the required locus is

\[{x^2} - 2y + 12 = 0\]