Basic Examples on Parabolas Set 1

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Example - 3

Find the length of the LR of the parabola \(4{y^2} + 12x - 12y + 39 = 0\)

Solution: The equation can be rearranged as (verify)

\[\begin{align}&\qquad \;\;{\left( {y - \frac{3}{2}} \right)^2} = - 3\left( {x + \frac{5}{2}} \right)\\&\Rightarrow \quad {\left( {y - \frac{3}{2}} \right)^2} = - 4 \times \left( {\frac{3}{4}} \right) \times \left( {x + \frac{5}{2}} \right)\end{align}\]

This is of the form

\[{Y^2} = - 4aX{\rm{ }}\left( {a = \frac{3}{4}} \right)\]

i.e. this is a parabola with vertex \(\left(\begin{align} {\frac{3}{2},\,\,\frac{5}{2}}\end{align} \right)\) and axis parallel to the x-axis. The parabola opens to the left. We are only concerned with the length of the LR which is simply 4a, i.e. 3 units.

Example – 4

The axis of a parabola is the line \({L_1}:3x + 4y - 4 = 0\) and the tangent to it at the vertex is \({L_2}:4x - 3y + 7 = 0.\) The LR is 4 units in length. Find the equation of the parabola.

Solution: Consider for a moment, the co-ordinate axes system formed by \({L_1}and{\rm{ }}{L_2}\).

If \({L_1}\;and\; {\rm{ }}{L_2}\) were truly the actual co-ordinate axes, the equation of the parabola would have been

\[\begin{align}&\qquad \;\;{y^2} = 4ax\\\\&\Rightarrow \quad P{A^2} = ({\rm{Length\;of\; }}LR) \times PB...\left( 1 \right)\end{align}\]

Now, even if we use some other co-ordinate system (here the \({L_1} - {\rm{ }}{L_2}\) system), the relation (1) will still hold since that only depends on lengths which are invariant with respect to the co-ordinate system chosen. Thus, simply applying (1) will give us the required equation :

\[\begin{align}& \qquad\;{\left( {\frac{{\left| {3x + 4y - 4} \right|}}{5}} \right)^2} = (4) \times \frac{{\left| {4x - 3y + 7} \right|}}{5}\\&\Rightarrow \quad {(3x + 4y - 4)^2} = \pm 20(4x - 3y + 7)\end{align}\]

Thus, as might have been expected, we’ll obtain two parabolas with the given property, one opening to the ‘left’ and one to the ‘right’ in the \({L_1} - {\rm{ }}{L_2}\) co-ordinate system.

Example - 5

Show that the locus of the middle points of all chords of the parabola \({y^2} = 4ax\) which pass through the origin, is another parabola.

Solution: Let OP be any such chord of the parabola, where O is the origin (0, 0) and P is the point (h, k) lying somewhere on the parabola. We then have

The mid-point of OP, say P' is clearly \(\left( \begin{align}{\frac{h}{2},\,\,\frac{k}{2}} \end{align}\right)\) .

Since \({k^2} = 4ah,\) we have


\[\begin{align}&\qquad {\left( {\frac{k}{2}} \right)^2} = 2a\left( {\frac{h}{2}} \right)\\\\&\Rightarrow \quad {y^2} = 2ax\qquad\qquad\qquad...\left( 1 \right)\end{align}\]

where we used (x, y) instead of \(\left(\begin{align} {\frac{h}{2},\,\,\frac{k}{2}} \end{align}\right)\) to specify the equation of the locus of P' in the conventional x – y form.

(1) shows that the required locus is another parabola, which has the same vertex and axis.

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Parabolas
grade 11 | Questions Set 1
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grade 11 | Questions Set 2
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grade 11 | Answers Set 2
Download practice questions along with solutions for FREE:
Parabolas
grade 11 | Questions Set 1
Parabolas
grade 11 | Answers Set 1
Parabolas
grade 11 | Questions Set 2
Parabolas
grade 11 | Answers Set 2
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