Question

A circle and a parabola intersect in four points; show that the algebraic sum of the ordinates of the four points is zero.
Show also that the line joining one pair of these four points and the line joining the other pair are equally inclined to the axis.

Answer 1

Hint: Consider the general equations of parabola and circle and combine to form the biquadratic equation and use the relation of roots to get the desired result.

Complete step-by-step answer:

In the equation given, we are given a parabola and a circle.

So, now let’s consider general equation of circle which is

${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0.......(i)$

and of parabola, ${{y}^{2}}=4ax........(ii)$

It is given that the circle and parabola intersects at four points.

So, if we solve the equations (i) and (ii) we can find it,

Now substituting ${{y}^{2}}=4ax$ as $x=\dfrac{{{y}^{2}}}{4a}$ in equation (i) we get,

$\begin{align}

  & {{\left( \dfrac{{{y}^{2}}}{4a} \right)}^{2}}+{{y}^{2}}+2g\left( \dfrac{{{y}^{2}}}{4a} \right)+2fy+c=0 \\

 & \dfrac{{{y}^{4}}}{16{{a}^{2}}}+{{y}^{2}}+g\dfrac{{{y}^{2}}}{2a}+2fy+c=0 \\

\end{align}$

Now this equation is a fourth degree equation in y.

Let the roots of the equation be ${{y}_{1}},{{y}_{2}},{{y}_{3}},{{y}_{4}}$ .

By the relation of roots, we know that the formula for sum of roots is $-\dfrac{\text{coefficient of }{{\text{y}}^{3}}}{\text{coefficient of }{{\text{y}}^{4}}}$

which can be represented as,

${{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}=-\dfrac{0}{\dfrac{1}{16{{a}^{2}}}}=0$

Hence the sum of ordinates of points is ‘0’ which is now proved.

Let the points intersections of parabola and circle P, Q, R, S be \[\left( {{x}_{1}},{{y}_{1}}

\right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right),\left( {{x}_{4}},{{y}_{4}}

\right)\] respectively.

Now these points also lie on parabola so it also satisfies conditions of parabola too so,

\[\begin{align}

  & {{y}_{1}}^{2}=4a{{x}_{1}}........(i) \\

 & {{y}_{2}}^{2}=4a{{x}_{2}}........(ii) \\

\end{align}\]

Subtraction (ii) from (i) we get,

${{y}_{1}}^{2}-{{y}_{2}}^{2}=4a\left( {{x}_{1}}-{{x}_{2}} \right)$

We will use the identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ ,

$\Rightarrow \left( {{y}_{1}}+{{y}_{2}} \right)({{y}_{1}}-{{y}_{2}})=4a\left( {{x}_{1}}-{{x}_{2}}

\right)$

Rearranging this we will get

$\Rightarrow \dfrac{\left( {{y}_{1}}-{{y}_{2}} \right)}{\left( {{x}_{1}}-{{x}_{2}} \right)}=\dfrac{4a}{\left( {{y}_{1}}+{{y}_{2}} \right)}$

Slope of line PQ = $\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}=\dfrac{4a}{{{y}_{1}}+{{y}_{2}}}$
Now considering the points $\left( {{x}_{3}},{{y}_{3}} \right),\left( {{x}_{4}},{{y}_{4}} \right)$ in parabola we get,

\[\begin{align}

  & {{y}_{3}}^{2}=4a{{x}_{3}}........(iii) \\

 & {{y}_{4}}^{2}=4a{{x}_{4}}........(iv) \\

\end{align}\]

Subtraction (iv) from (iii) we get,

$\begin{align}

  & {{y}_{3}}^{2}-{{y}_{4}}^{2}=4a\left( {{x}_{3}}-{{x}_{4}} \right) \\

 & \Rightarrow \left( {{y}_{3}}+{{y}_{4}} \right)\left( {{y}_{3}}-{{y}_{4}} \right)=4a\left( {{x}_{3}}-{{x}_{4}} \right) \\

 & \Rightarrow \dfrac{\left( {{y}_{3}}-{{y}_{4}} \right)}{\left( {{x}_{3}}-{{x}_{4}} \right)}=\dfrac{4a}{{{y}_{3}}+{{y}_{4}}} \\

\end{align}$

Slope of line RS $=\dfrac{{{y}_{3}}-{{y}_{4}}}{{{x}_{3}}-{{x}_{4}}}=\dfrac{4a}{{{y}_{3}}+{{y}_{4}}}$
As we know that $\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}} \right)=0$ it can be written as,
${{y}_{1}}+{{y}_{2}}=-\left( {{y}_{3}}+{{y}_{4}} \right)$

So, Slope of line RS = $\dfrac{4a}{-\left( {{y}_{1}}+{{y}_{2}} \right)}=-\dfrac{4a}{{{y}_{1}}+{{y}_{2}}}$

We can write it as,

(Slope of line RS) = – (Slope of line PQ)

So, the slopes of RS and PQ are equal in magnitude; hence PQ and RS are equally inclined to the axis.


Note: Students shall be careful while combining the general equation of circle and parabola to make a single biquadratic equation.