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# Parabolas Set 3

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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

${x^2} - 2y + 12 = 0$

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