PR and QR are chords of a parabola which are normal at P and Q. prove that two of the common chords of the parabola and the circle circumscribing the triangle PRQ meet on the directrix.
Answer
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Hint: Draw a suitable diagram with the help of given information in the problem to visualize the problem. Normal at any point $\left( a{{t}^{2}},2at \right)$ for the parabola ${{y}^{2}}=4ax$ is given as
$y=-tx+2at+a{{t}^{3}}$
Complete step-by-step answer:
Suppose three points P, Q, R as parametric form of coordinates using three different variables. As the equation of normal (mentioned above) is cubic in ‘t’, so try to get equations with respect to roots of the cubic; which has roots same as the variables supposed for representing P, Q, R. suppose another point S on the parabola, which is fourth intersect point of circle and parabola. Write equations of RS and PQ and find their intersects to verify whether that point is lying on the directrix of ${{y}^{2}}=4ax$. Directrix of ${{y}^{2}}=4ax$ is given as x + a = 0, x = -a.
As PR and QR are the chords of a parabola, where PR and QR are normal at P and Q respectively, and meeting R which is lying on the parabola as well.
Now, we need to prove two of the common chords of the parabola and circle circumscribing the triangle PQR meets on the directrix.
So, let us draw the diagram with the help of given information as
Let the coordinates of R be (h, k). And let us suppose the coordinates of R in parametric form be $\left( a{{t}^{2}}_{1},2a{{t}_{1}} \right)$
So, we get
\[h=a{{t}^{2}},\]k = 2at ……………. (i)
Again, let the coordinates of P and Q are
$\left( a{{t}^{2}}_{1},2a{{t}_{1}} \right),\left( a{{t}^{2}}_{2},2a{{t}_{2}} \right)$ respectively.
Now, we know the equation of normal at any point $\left( a{{t}^{2}},2at \right)$ for the parabola${{y}^{2}}=4ax$ is given as
$y=-tx+2at+a{{t}^{3}}$ …………….(ii)
Now, if the normal given in equation (ii) will pass through (h, k), then we have
$\begin{align}
& y=-tx+2at+a{{t}^{3}}, \\
& a{{t}^{3}}+t\left( 2a-x \right)-y=0 \\
\end{align}$
As, the above equation is passing through R(h, k), then, we get
$a{{t}^{3}}+t\left( 2a-h \right)-k=0$ …………….(iii)
Now, the above equation is cubic in terms of ‘t’, so, there must be three roots, $t,{{t}_{1}},{{t}_{2}}$ which represents the points P, Q, R. so, we get sum of roots of equation (iii) = 0
${{t}_{1}}+{{t}_{2}}+t=0$ …………(iv)
${{t}_{1}}{{t}_{2}}+{{t}_{2}}t+t{{t}_{1}}=\dfrac{2a-h}{a}$ ………..(v)
$t{{t}_{1}}{{t}_{2}}=\dfrac{k}{a}$ ……………..(vi)
But as we supposed
$h=a{{t}^{2}},$ k = 2at
So, we can eliminate ‘t’ from the above equations calculated from cubic equation, with the help of
k = 2at $\to t=\dfrac{k}{2a}$
so, we get equation (iv) as
${{t}_{1}}+{{t}_{2}}+\dfrac{k}{2a}=0$
${{t}_{1}}+{{t}_{2}}=-\dfrac{k}{2a}$………………….(vii)
From equation (vi)
$\begin{align}
& {{t}_{1}}{{t}_{2}}t=\dfrac{k}{2a} \\
& {{t}_{1}}{{t}_{2}}\times \dfrac{k}{2a}=\dfrac{k}{a} \\
\end{align}$
${{t}_{1}}{{t}_{2}}=2$……………(viii)
Now, equations (vii) and (viii) suggest that the circle and parabola intersect in four points and the line joining the other pair is equally inclined with the axis. Therefore, if ‘s’ be the fourth point, then PQ and RS are equally inclined with the axis.
As ${{t}_{1}}{{t}_{2}}=2,$ so the line joining the pairs of four points will be equally inclined to the axis.
Now, the equation of PQ can be calculated by the general equation of line
$\begin{align}
& y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) \\
& \left( a{{t}^{2}}_{1},2a{{t}_{2}} \right),\left( a{{t}^{2}}_{1},2a{{t}_{1}} \right) \\
\end{align}$
So, we get equation of PQ as
$\begin{align}
& y-2a{{t}_{2}}=\dfrac{2a{{t}_{2}}-2a{{t}_{1}}}{a{{t}_{2}}^{2}-a{{t}_{1}}^{2}}\left( x-a{{t}_{2}}^{2} \right) \\
& y-2a{{t}_{2}}=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}^{2}-{{t}_{1}}^{2} \right)}\left( x-a{{t}_{2}}^{2} \right) \\
& y-2a{{t}_{2}}=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{1}}+{{t}_{2}} \right)\left( {{t}_{2}}-{{t}_{1}} \right)}\left( x-a{{t}_{2}}^{2} \right) \\
& y-2a{{t}_{2}}=\dfrac{2}{{{t}_{1}}+{{t}_{2}}}\left( x-a{{t}_{2}}^{2} \right) \\
& \left( y-2a{{t}_{2}} \right)\left( {{t}_{1}}+{{t}_{2}} \right)=2x-2a{{t}_{2}}^{2} \\
& y{{t}_{1}}+y{{t}_{2}}-2a{{t}_{1}}{{t}_{2}}-2a{{t}_{2}}^{2}=2x-2a{{t}_{2}}^{2} \\
& y\left( {{t}_{1}}+{{t}_{2}} \right)-2a{{t}_{1}}{{t}_{2}}=2x \\
\end{align}$
From equation (vii) and (viii), we get the above equation as
$\begin{align}
& y\times \dfrac{-k}{2a}-2a\times 2=2x \\
& \dfrac{-ky}{2a}-4a=2x \\
& -ky-8{{a}^{2}}=4ax \\
\end{align}$
$-ky=4ax+8{{a}^{2}}$………………(ix)
Similarly, the equation of RS can be given as
As PQ and RS are equally inclined to x-axis (axis of the parabola), so, slope of RS will be $\dfrac{4a}{k}$ as well [slope of PQ is $\dfrac{-4a}{k}$ from equation (ix) and both cannot be equal].
And R has the coordinates (h, k).
So, we get equation of RS as
$\begin{align}
& y-k=\dfrac{4a}{k}\left( x-h \right) \\
& k\left( y-k \right)=4ax-4ah \\
& ky-{{k}^{2}}=4ax-4ah \\
& ky=4ax+{{k}^{2}}-4ah \\
\end{align}$
Now, as (h, k) is passing through the parabola ${{y}^{2}}=4ax$. So, we get
${{k}^{2}}=4ah$
Hence, we get equation of RS as
ky = 4ax ……………(x)
adding equation (ix) and (x), we get
$\begin{align}
& -ky+ky=4ax+8{{a}^{2}}+4ax \\
& 0=8ax+8{{a}^{2}} \\
& 8a\left( x+a \right)=0 \\
& x+a=0 \\
\end{align}$
Where, $a\ne 0$
x = -a
which is the equation of directrix. Hence, the common chords and RS of the parabola and circle meet in the directrix.
Note: We can find the equation of normal for any parabola ${{y}^{2}}=4ax$ at point $\left( a{{t}^{2}},2at \right)$ by finding the slope of tangent with the help of differentiation at that point and hence, we can get slope of normal as well (product of slopes of two perpendicular lines =-1). Now, get the equation of normal using the point-slope form of a line $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. So, it can be an approach to get the equation of normal, using direct results in conics always makes the problem easier and flexible, so, try to remember the equation of tangents and normal to any conic to solve this kind of problems in less time.
One may suppose the points P, Q, R as
$\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ as well, but it will the problem complex and will take longer time.so, try to use parametric coordinates to solve this kind of problems.
As PQ and RS are different lines and it is given that their inclination is the same with the axis. It means one line will have a slope of $m\left( \tan \theta \right)$ and another should have $-m\left( -\tan \theta =\tan \left( -\theta \right) \right)$. Hence, be careful with this step in the problem.
$y=-tx+2at+a{{t}^{3}}$
Complete step-by-step answer:
Suppose three points P, Q, R as parametric form of coordinates using three different variables. As the equation of normal (mentioned above) is cubic in ‘t’, so try to get equations with respect to roots of the cubic; which has roots same as the variables supposed for representing P, Q, R. suppose another point S on the parabola, which is fourth intersect point of circle and parabola. Write equations of RS and PQ and find their intersects to verify whether that point is lying on the directrix of ${{y}^{2}}=4ax$. Directrix of ${{y}^{2}}=4ax$ is given as x + a = 0, x = -a.
As PR and QR are the chords of a parabola, where PR and QR are normal at P and Q respectively, and meeting R which is lying on the parabola as well.
Now, we need to prove two of the common chords of the parabola and circle circumscribing the triangle PQR meets on the directrix.
So, let us draw the diagram with the help of given information as
Let the coordinates of R be (h, k). And let us suppose the coordinates of R in parametric form be $\left( a{{t}^{2}}_{1},2a{{t}_{1}} \right)$
So, we get
\[h=a{{t}^{2}},\]k = 2at ……………. (i)
Again, let the coordinates of P and Q are
$\left( a{{t}^{2}}_{1},2a{{t}_{1}} \right),\left( a{{t}^{2}}_{2},2a{{t}_{2}} \right)$ respectively.
Now, we know the equation of normal at any point $\left( a{{t}^{2}},2at \right)$ for the parabola${{y}^{2}}=4ax$ is given as
$y=-tx+2at+a{{t}^{3}}$ …………….(ii)
Now, if the normal given in equation (ii) will pass through (h, k), then we have
$\begin{align}
& y=-tx+2at+a{{t}^{3}}, \\
& a{{t}^{3}}+t\left( 2a-x \right)-y=0 \\
\end{align}$
As, the above equation is passing through R(h, k), then, we get
$a{{t}^{3}}+t\left( 2a-h \right)-k=0$ …………….(iii)
Now, the above equation is cubic in terms of ‘t’, so, there must be three roots, $t,{{t}_{1}},{{t}_{2}}$ which represents the points P, Q, R. so, we get sum of roots of equation (iii) = 0
${{t}_{1}}+{{t}_{2}}+t=0$ …………(iv)
${{t}_{1}}{{t}_{2}}+{{t}_{2}}t+t{{t}_{1}}=\dfrac{2a-h}{a}$ ………..(v)
$t{{t}_{1}}{{t}_{2}}=\dfrac{k}{a}$ ……………..(vi)
But as we supposed
$h=a{{t}^{2}},$ k = 2at
So, we can eliminate ‘t’ from the above equations calculated from cubic equation, with the help of
k = 2at $\to t=\dfrac{k}{2a}$
so, we get equation (iv) as
${{t}_{1}}+{{t}_{2}}+\dfrac{k}{2a}=0$
${{t}_{1}}+{{t}_{2}}=-\dfrac{k}{2a}$………………….(vii)
From equation (vi)
$\begin{align}
& {{t}_{1}}{{t}_{2}}t=\dfrac{k}{2a} \\
& {{t}_{1}}{{t}_{2}}\times \dfrac{k}{2a}=\dfrac{k}{a} \\
\end{align}$
${{t}_{1}}{{t}_{2}}=2$……………(viii)
Now, equations (vii) and (viii) suggest that the circle and parabola intersect in four points and the line joining the other pair is equally inclined with the axis. Therefore, if ‘s’ be the fourth point, then PQ and RS are equally inclined with the axis.
As ${{t}_{1}}{{t}_{2}}=2,$ so the line joining the pairs of four points will be equally inclined to the axis.
Now, the equation of PQ can be calculated by the general equation of line
$\begin{align}
& y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) \\
& \left( a{{t}^{2}}_{1},2a{{t}_{2}} \right),\left( a{{t}^{2}}_{1},2a{{t}_{1}} \right) \\
\end{align}$
So, we get equation of PQ as
$\begin{align}
& y-2a{{t}_{2}}=\dfrac{2a{{t}_{2}}-2a{{t}_{1}}}{a{{t}_{2}}^{2}-a{{t}_{1}}^{2}}\left( x-a{{t}_{2}}^{2} \right) \\
& y-2a{{t}_{2}}=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}^{2}-{{t}_{1}}^{2} \right)}\left( x-a{{t}_{2}}^{2} \right) \\
& y-2a{{t}_{2}}=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{1}}+{{t}_{2}} \right)\left( {{t}_{2}}-{{t}_{1}} \right)}\left( x-a{{t}_{2}}^{2} \right) \\
& y-2a{{t}_{2}}=\dfrac{2}{{{t}_{1}}+{{t}_{2}}}\left( x-a{{t}_{2}}^{2} \right) \\
& \left( y-2a{{t}_{2}} \right)\left( {{t}_{1}}+{{t}_{2}} \right)=2x-2a{{t}_{2}}^{2} \\
& y{{t}_{1}}+y{{t}_{2}}-2a{{t}_{1}}{{t}_{2}}-2a{{t}_{2}}^{2}=2x-2a{{t}_{2}}^{2} \\
& y\left( {{t}_{1}}+{{t}_{2}} \right)-2a{{t}_{1}}{{t}_{2}}=2x \\
\end{align}$
From equation (vii) and (viii), we get the above equation as
$\begin{align}
& y\times \dfrac{-k}{2a}-2a\times 2=2x \\
& \dfrac{-ky}{2a}-4a=2x \\
& -ky-8{{a}^{2}}=4ax \\
\end{align}$
$-ky=4ax+8{{a}^{2}}$………………(ix)
Similarly, the equation of RS can be given as
As PQ and RS are equally inclined to x-axis (axis of the parabola), so, slope of RS will be $\dfrac{4a}{k}$ as well [slope of PQ is $\dfrac{-4a}{k}$ from equation (ix) and both cannot be equal].
And R has the coordinates (h, k).
So, we get equation of RS as
$\begin{align}
& y-k=\dfrac{4a}{k}\left( x-h \right) \\
& k\left( y-k \right)=4ax-4ah \\
& ky-{{k}^{2}}=4ax-4ah \\
& ky=4ax+{{k}^{2}}-4ah \\
\end{align}$
Now, as (h, k) is passing through the parabola ${{y}^{2}}=4ax$. So, we get
${{k}^{2}}=4ah$
Hence, we get equation of RS as
ky = 4ax ……………(x)
adding equation (ix) and (x), we get
$\begin{align}
& -ky+ky=4ax+8{{a}^{2}}+4ax \\
& 0=8ax+8{{a}^{2}} \\
& 8a\left( x+a \right)=0 \\
& x+a=0 \\
\end{align}$
Where, $a\ne 0$
x = -a
which is the equation of directrix. Hence, the common chords and RS of the parabola and circle meet in the directrix.
Note: We can find the equation of normal for any parabola ${{y}^{2}}=4ax$ at point $\left( a{{t}^{2}},2at \right)$ by finding the slope of tangent with the help of differentiation at that point and hence, we can get slope of normal as well (product of slopes of two perpendicular lines =-1). Now, get the equation of normal using the point-slope form of a line $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. So, it can be an approach to get the equation of normal, using direct results in conics always makes the problem easier and flexible, so, try to remember the equation of tangents and normal to any conic to solve this kind of problems in less time.
One may suppose the points P, Q, R as
$\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ as well, but it will the problem complex and will take longer time.so, try to use parametric coordinates to solve this kind of problems.
As PQ and RS are different lines and it is given that their inclination is the same with the axis. It means one line will have a slope of $m\left( \tan \theta \right)$ and another should have $-m\left( -\tan \theta =\tan \left( -\theta \right) \right)$. Hence, be careful with this step in the problem.
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