Question

A quadrilateral is inscribed in a parabola ${{y}^{2}}=4ax$ and three of its sides pass through fixed points on the axis. Show that the fourth side also passes through a fixed point on the axis of the parabola.

Answer 1

Hint: Draw figure as mentioned. Take quadrilateral as ABCD. Consider the coordinate of A as $\left( at_{1}^{2},2{{t}_{1}} \right)$ . Fixed slope of AB, thus get the equation of line AB. Take A, B, C, D as ${{t}_{1}},{{t}_{2}},{{t}_{3}},{{t}_{4}}$ . From the equation of line, you get ${{t}_{1}},{{t}_{2}}$ . similarly find ${{t}_{2}},{{t}_{3}},{{t}_{4}}$ and multiply them together to find ${{t}_{1}}-{{t}_{4}}$ as constant.

Complete step by step solution:

We have been given this equation of parabola as ${{y}^{2}}=4ax$ . It is said that a quadrilateral is inscribed in this parabola. The 3 sides of the quadrilateral passes through fixed points in the axis. We know that a quadrilateral has 4 sides. As 3 sides are fixed, we need to show that the 4th side of the quadrilateral is also fixed on the axis of the parabola. Let us take the quadrilateral ABCD.

Point A and B touches the parabola ${{y}^{2}}=4ax$. Let us take the coordinates ${{t}_{1}},{{t}_{2}},{{t}_{3}}$ and ${{t}_{4}}$ for A,B, C and D.

Put $x=at_{1}^{2}$ in the equation of parabola. ${{y}^{2}}=4ax$

$\Rightarrow {{y}^{2}}=4a\left( at_{1}^{2} \right)={{y}^{2}}=4{{a}^{2}}t_{1}^{2}$ .

Take square of both sides,

$y=2a{{t}_{1}}$ . Thus, the coordinates of A are $\left( x,y \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)$ .

Similarly, the coordinates of B become $\left( at_{2}^{2},2a{{t}_{2}} \right)$ .

Now let us find the slope of line AB. We know the formula for finding slope,

$\text{slope=}\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ here, $\left( {{x}_{1}},{{y}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)$

$\therefore$ slope of AB $=\dfrac{2a{{t}_{2}}-2a{{t}_{1}}}{at_{2}^{2}-at_{1}^{2}}=\dfrac{2a\left( {{t}_{2}}-{{t}_{1}} \right)}{a\left( t_{2}^{2}-t_{1}^{2} \right)}$ .

We know that $\left( t_{2}^{2}-t_{1}^{2} \right)=\left( {{t}_{2}}+{{t}_{1}} \right)\left( {{t}_{2}}-{{t}_{1}} \right)$ i.e. similar to ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .

$\therefore$ Slope of AB $=\dfrac{2a\left( {{t}_{2}}-{{t}_{1}} \right)}{a\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)}$ .

Cancel out $a\left( {{t}_{2}}-{{t}_{1}} \right)$ from the numerator & denominator.

$\therefore$ Slope of AB $=\dfrac{2}{{{t}_{2}}+{{t}_{1}}}=m$

Thus, we can find the equation of line AB by using the formula, $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ .

Put $\left( {{x}_{1}},{{y}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)$ and $m=\dfrac{2}{{{t}_{2}}+{{t}_{1}}}$ .

$y-2a{{t}_{1}}=\dfrac{2}{{{t}_{2}}+{{t}_{1}}}\left( x-at_{1}^{2} \right)$

Now let us simplify the above expression,

$\begin{aligned}

& \left( {{t}{1}}+{{t}{2}} \right)\left[ y-2a{{t}{1}} \right]=2\left( x-at{1}^{2} \right) \\

& y\left( {{t}{1}}+{{t}{2}} \right)-2a{{t}{1}}\left( {{t}{1}}+{{t}{2}} \right)=2x-2at{1}^{2} \\

& y\left( {{t}{1}}+{{t}{2}} \right)-2at_{1}^{2}-2a{{t}{1}}{{t}{2}}=2x-2at_{1}^{2} \\

\end{aligned}$

Cancel out $2at_{1}^{2}$ from LHS & RHS

$y\left( {{t}_{1}}+{{t}_{2}} \right)-2a{{t}_{1}}{{t}_{2}}=2x$

Now this line passes through a fixed point on the axis. So, we can put y = 0.

Putting the value, we get

$\begin{aligned}

& 0\times \left( {{t}{1}}+{{t}{2}} \right)-2a{{t}{1}}{{t}{2}}=2x \\

& \Rightarrow x=-a{{t}{1}}{{t}{2}} \\

\end{aligned}$

Let us take $-a{{t}_{1}}{{t}_{2}}=x={{k}_{1}}$ .

$\therefore {{k}_{1}}=-a{{t}_{1}}{{t}_{2}}$ , ${{t}_{1}}{{t}_{2}}=-\dfrac{{{k}_{1}}}{a}$

From equation of chord AB, we got ${{t}_{1}}{{t}_{2}}=-\dfrac{{{k}_{1}}}{a}$ ……………… (2)

Similarly for BC, we get ${{t}_{2}}{{t}_{3}}=-\dfrac{{{k}_{2}}}{a}$ …………….(3)

Similarly, for CD, we get ${{t}_{3}}{{t}_{4}}=-\dfrac{{{k}_{3}}}{a}$ ………………… (4)

We get this because it’s told that 3 sides through fixed point and A, B, C, D are taken as ${{t}_{1}},{{t}_{2}},{{t}_{3}},{{t}_{4}}$ .

Now let us multiply (2), (3) and (4), we get,

$\left( {{t}_{1}}{{t}_{2}} \right)\times \left( {{t}_{2}}{{t}_{3}} \right)\times \left( {{t}_{3}}{{t}_{4}} \right)=\left( -\dfrac{{{k}_{1}}}{a} \right)\left( -\dfrac{{{k}_{2}}}{a} \right)\left( -\dfrac{{{k}_{3}}}{a} \right)$

Let us simplify the above expression.

$\begin{aligned}

& {{t}{1}}t{2}^{2}t_{3}^{2}{{t}{4}}=-\dfrac{{{k}{1}}{{k}{2}}{{k}{3}}}{{{a}^{3}}} \\

& \Rightarrow {{t}{1}}{{t}{4}}\left( t_{2}^{2}t_{3}^{2} \right)=-\dfrac{{{k}{1}}{{k}{2}}{{k}_{3}}}{{{a}^{3}}} \\

\end{aligned}$

Put ${{t}_{2}}{{t}_{3}}=-\dfrac{{{k}_{2}}}{a}$

& {{t}_{1}}{{t}_{4}}{{\left( {{t}_{2}}{{t}_{3}} \right)}^{2}}=-\dfrac{{{k}_{1}}{{k}_{2}}{{k}_{3}}}{{{a}^{3}}} \\\ & \Rightarrow {{t}_{1}}{{t}_{4}}{{\left( -\dfrac{{{k}_{2}}}{a} \right)}^{2}}=-\dfrac{{{k}_{1}}{{k}_{2}}{{k}_{3}}}{{{a}^{3}}} \\\ & \Rightarrow {{t}_{1}}{{t}_{4}}\left( \dfrac{k_{2}^{2}}{{{a}^{2}}} \right)=-\dfrac{{{k}_{1}}{{k}_{2}}{{k}_{3}}}{{{a}^{3}}} \\\ \end{aligned}$$ Cancel out $\dfrac{{{k}_{2}}}{{{a}^{2}}}$ from both sides. $${{t}_{1}}{{t}_{4}}\times {{k}_{2}}=-\dfrac{{{k}_{1}}{{k}_{3}}}{a}$$ $\therefore {{t}_{1}}{{t}_{4}}=-\dfrac{{{k}_{1}}{{k}_{3}}}{a{{k}_{2}}}=\text{constant}$ Hence, from this we can say that the 4th side of the quadrilateral also passes through a fixed point on the axis of parabola. Note: Parabola is not one of the easy concepts. It is important that you consider the 6 sides of the quadrilateral as ${{t}_{1}},{{t}_{2}},{{t}_{3}},{{t}_{4}}$ . Thus, as 3 points are fixed which are ${{t}_{1}}{{t}_{2}},{{t}_{2}}{{t}_{3}}$ and ${{t}_{3}}{{t}_{4}}$ , you need to find this ${{t}_{3}}{{t}_{4}}$ is constant and hence fixed.