A variable circle passes through the point A $\left( 2,1 \right)$ and touches the x axis. Locus of the other end of the diameter through A is parabola.
The two tangents and two normal at the extremities of the latus rectum of the parabola constitute a quadrilateral. Find the area of the quadrilateral.
Answer
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Hint: First find the equation of parabola since the point on the other end of diameter A $\left( 2,1 \right)$ of a circle lies on the parabola and determine the points of the quadrilateral vertex to get its area. And draw the quadrilateral on a graph and carefully observe the graph formation before calculating the area.
Complete step by step answer:
It is given in the question that A $\left( 2,1 \right)$ lies on the circle. And the other end of its diameter lies on the parabola. So, let B $(p,q)$ be the other endpoint which lies on the parabola. So, AB is the diameter of the circle.
We know the equation of a circle in diameter form when two endpoints of diameter is $({{x}_{1}},{{y}_{1}})\And ({{x}_{2}},{{y}_{2}})$, then, the equation is in the form $(x-{{x}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})=0$.
Now, equation of the circle in diameter form is:
$\left( x-p \right)\left( x-2 \right)+\left( y-q \right)\left( y-1 \right)=0$
As it is given in the question that circle touches the x axis so y coordinate is equal to zero,
$\left( x-p \right)\left( x-2 \right)+(-q)(-1)=0$
$\begin{align}
& {{x}^{2}}-2x-px+2p+q=0 \\
& {{x}^{2}}+(-2-p)x+2p+q=0 \\
\end{align}$
It forms a quadratic equation such as $a{{x}^{2}}+bx+c=0$, the discriminant is $d={{b}^{2}}-4ac$. Since the circle cuts x-axis, the discriminant d is equal to 0. So, we get
$\begin{align}
& {{(-2-p)}^{2}}-4(2p+q)=0 \\
& 4+{{p}^{2}}+4p-8p-4q=0 \\
& {{p}^{2}}-4p+4-4q=0 \\
\end{align}$
Using completing the square method,
$\begin{align}
& {{(p-2)}^{2}}-4+4-4q=0 \\
& {{(p-2)}^{2}}=4q \\
\end{align}$
Replacing $(p,q)$ with $\left( x,y \right)$ we get,
${{(x-2)}^{2}}=4y$ ..(1)
We know the standard form of parabola is ${{(x-h)}^{2}}=4a(y-k)$, where focus is at $(h,k+a)$ and vertex is at $(h,k)$.
So comparing the equation we get ${{(x-2)}^{2}}=4y$ with the standard form of parabola ${{(x-h)}^{2}}=4a(y-k)$. We get the vertex of parabola at $(2,0)$ and focus at $\left( 2,1 \right)$. We get the graph of parabola as shown below.
Now we know we need to draw two tangents and two normal at the extremities of the latus rectum of the parabola. So, we know that here latus rectum is parallel to the x-axis (since latus rectum is parallel to the directrix of parabola). Therefore y coordinate of the point at the latus rectum is always the same, so here $y=1$.
Now, find the extremities point of the latus rectum. Putting value of $y=1$ in equation (1) we get,
${{(x-2)}^{2}}=4(1)$
Taking square root on both sides, we get
$(x-2)=\pm 2$
Now, let us consider each value as,
$\begin{align}
& \left( x-2 \right)=2 \\
& x=4 \\
\end{align}$
Then one of the points is B $\left( 4,1 \right)$.
$\begin{align}
& (x-2)=-2 \\
& x=0 \\
\end{align}$
Then another point is C $(0,1)$.
Now we get the points through which we need to draw tangent and normal. So now consider point B $\left( 4,1 \right)$, draw tangent and normal passing through point B. First we need to find the slope of tangent and normal.
Slope of the tangent $m1=\dfrac{dy}{dx}$
${{(x-2)}^{2}}=4y$
We have to differentiate w.r.t. x on both sides and we know that $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ , so
$2(x-2)=4\dfrac{dy}{dx}$
Now on solving,
${{m}_{1}}=\dfrac{(x-2)}{2}$
Slope of tangent 1 at point $\left( 4,1 \right)$ is ${{m}_{1}}=1$.
Now we know that the tangent and normal to a curve through the same point are perpendicular to each other. So the product of slope of two perpendicular lines is equal to $-1$. Therefore let the slope of normal be ${{n}_{1}}$. We have ${{m}_{1}}{{n}_{1}}=-1$ , as ${{m}_{1}}=1$ then ${{n}_{1}}=-1$.
Equation of tangent 1 passing through $\left( 4,1 \right)$: $y-{{y}_{1}}={{m}_{1}}(x-{{x}_{1}})$. So, we get
$y-1=1(x-4)$
$x-y-3=0$ …(2)
Equation of normal 1 passing through $\left( 4,1 \right)$
$\begin{align}
& y-{{y}_{1}}={{n}_{1}}(x-{{x}_{1}}) \\
& y-1=-1(x-4) \\
\end{align}$
So, we get
$x+y-5=0$ …(3)
Now similarly we find the equation of tangent and normal through point C $(0,1)$. Let slope of tangent 2 through point C $(0,1)$ be ${{m}_{2}}$ and the slope of normal 2 through point C $(0,1)$ be ${{n}_{2}}$. Then, we know that $2(x-2)=4\dfrac{dy}{dx}$, so we get, ${{m}_{2}}=\dfrac{(x-2)}{2}$.
The slope of the tangent at point C $(0,1)$ is ${{m}_{2}}=-1$ and hence the slope of normal at point C $(0,1)$ is ${{n}_{2}}=1$.
Now equation of tangent 2 passing through C $(0,1)$ is
$\begin{align}
& y-1=-1(x-0) \\
& y-1=-x \\
\end{align}$
$x+y-1=0$ …(4)
Now the equation of normal 2 passing through C $(0,1)$ is
$\begin{align}
& y-1=1(x-0) \\
& y-1=x \\
\end{align}$
$x-y+1=0$ ..(5)
Now we have the equation of both tangent and normal to find the intersection point between tangent and normal respectively.
Solving equation (2) & (4) we get the point of intersection of tangent. Adding (2) & (4), we get $x=2$. Now putting the value of x in equation (2), we get $y=-1$. Let us name that point D $\left( 2,-1 \right)$.
Now solving equation (3) & (5) we get the intersection point of normal. Adding (3) & (5), we get $x=2$. And putting the value of y in equation (3), we get, $y=3$. Let us name that point E $\left( 2,3 \right)$.
Quadrilateral forms having coordinates B $\left( 4,1 \right)$, C $(0,1)$, D $\left( 2,-1 \right)$and E $\left( 2,3 \right)$. Observing the diagram on graph carefully, we know that tangent and normal are perpendicular to each other. So CD is perpendicular to the EC and BD is perpendicular to the EB. We know the distance formula, distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right)$ is equal to$\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$.
So CD $=2\sqrt{2}$
EC $=2\sqrt{2}$
BD $=2\sqrt{2}$
EB $=2\sqrt{2}$
ED $=4$
BC $=4$
So all sides are equal to quadrilateral and each angle makes ${{90}^{\circ }}$ and diagonal are also equal, so the resulting quadrilateral will be a square.
Area of square (quadrilateral) $=sid{{e}^{2}}$
Area of square (quadrilateral) $={{\left( 2\sqrt{2} \right)}^{2}}$ $=8$ sq. units.
Note: We can use an alternate method to determine the area of quadrilateral by dividing quadrilateral into two triangles and area of triangle can be found using coordinate geometry formula when the vertices of triangle points are given such that, for example when three vertices points are given such as $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right)$, then area is equal to $\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)$. Using that formula we can find the area of two triangles and then the area of the quadrilateral is the sum of the area of those two triangles.
Complete step by step answer:
It is given in the question that A $\left( 2,1 \right)$ lies on the circle. And the other end of its diameter lies on the parabola. So, let B $(p,q)$ be the other endpoint which lies on the parabola. So, AB is the diameter of the circle.
We know the equation of a circle in diameter form when two endpoints of diameter is $({{x}_{1}},{{y}_{1}})\And ({{x}_{2}},{{y}_{2}})$, then, the equation is in the form $(x-{{x}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})=0$.
Now, equation of the circle in diameter form is:
$\left( x-p \right)\left( x-2 \right)+\left( y-q \right)\left( y-1 \right)=0$
As it is given in the question that circle touches the x axis so y coordinate is equal to zero,
$\left( x-p \right)\left( x-2 \right)+(-q)(-1)=0$
$\begin{align}
& {{x}^{2}}-2x-px+2p+q=0 \\
& {{x}^{2}}+(-2-p)x+2p+q=0 \\
\end{align}$
It forms a quadratic equation such as $a{{x}^{2}}+bx+c=0$, the discriminant is $d={{b}^{2}}-4ac$. Since the circle cuts x-axis, the discriminant d is equal to 0. So, we get
$\begin{align}
& {{(-2-p)}^{2}}-4(2p+q)=0 \\
& 4+{{p}^{2}}+4p-8p-4q=0 \\
& {{p}^{2}}-4p+4-4q=0 \\
\end{align}$
Using completing the square method,
$\begin{align}
& {{(p-2)}^{2}}-4+4-4q=0 \\
& {{(p-2)}^{2}}=4q \\
\end{align}$
Replacing $(p,q)$ with $\left( x,y \right)$ we get,
${{(x-2)}^{2}}=4y$ ..(1)
We know the standard form of parabola is ${{(x-h)}^{2}}=4a(y-k)$, where focus is at $(h,k+a)$ and vertex is at $(h,k)$.
So comparing the equation we get ${{(x-2)}^{2}}=4y$ with the standard form of parabola ${{(x-h)}^{2}}=4a(y-k)$. We get the vertex of parabola at $(2,0)$ and focus at $\left( 2,1 \right)$. We get the graph of parabola as shown below.
Now we know we need to draw two tangents and two normal at the extremities of the latus rectum of the parabola. So, we know that here latus rectum is parallel to the x-axis (since latus rectum is parallel to the directrix of parabola). Therefore y coordinate of the point at the latus rectum is always the same, so here $y=1$.
Now, find the extremities point of the latus rectum. Putting value of $y=1$ in equation (1) we get,
${{(x-2)}^{2}}=4(1)$
Taking square root on both sides, we get
$(x-2)=\pm 2$
Now, let us consider each value as,
$\begin{align}
& \left( x-2 \right)=2 \\
& x=4 \\
\end{align}$
Then one of the points is B $\left( 4,1 \right)$.
$\begin{align}
& (x-2)=-2 \\
& x=0 \\
\end{align}$
Then another point is C $(0,1)$.
Now we get the points through which we need to draw tangent and normal. So now consider point B $\left( 4,1 \right)$, draw tangent and normal passing through point B. First we need to find the slope of tangent and normal.
Slope of the tangent $m1=\dfrac{dy}{dx}$
${{(x-2)}^{2}}=4y$
We have to differentiate w.r.t. x on both sides and we know that $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ , so
$2(x-2)=4\dfrac{dy}{dx}$
Now on solving,
${{m}_{1}}=\dfrac{(x-2)}{2}$
Slope of tangent 1 at point $\left( 4,1 \right)$ is ${{m}_{1}}=1$.
Now we know that the tangent and normal to a curve through the same point are perpendicular to each other. So the product of slope of two perpendicular lines is equal to $-1$. Therefore let the slope of normal be ${{n}_{1}}$. We have ${{m}_{1}}{{n}_{1}}=-1$ , as ${{m}_{1}}=1$ then ${{n}_{1}}=-1$.
Equation of tangent 1 passing through $\left( 4,1 \right)$: $y-{{y}_{1}}={{m}_{1}}(x-{{x}_{1}})$. So, we get
$y-1=1(x-4)$
$x-y-3=0$ …(2)
Equation of normal 1 passing through $\left( 4,1 \right)$
$\begin{align}
& y-{{y}_{1}}={{n}_{1}}(x-{{x}_{1}}) \\
& y-1=-1(x-4) \\
\end{align}$
So, we get
$x+y-5=0$ …(3)
Now similarly we find the equation of tangent and normal through point C $(0,1)$. Let slope of tangent 2 through point C $(0,1)$ be ${{m}_{2}}$ and the slope of normal 2 through point C $(0,1)$ be ${{n}_{2}}$. Then, we know that $2(x-2)=4\dfrac{dy}{dx}$, so we get, ${{m}_{2}}=\dfrac{(x-2)}{2}$.
The slope of the tangent at point C $(0,1)$ is ${{m}_{2}}=-1$ and hence the slope of normal at point C $(0,1)$ is ${{n}_{2}}=1$.
Now equation of tangent 2 passing through C $(0,1)$ is
$\begin{align}
& y-1=-1(x-0) \\
& y-1=-x \\
\end{align}$
$x+y-1=0$ …(4)
Now the equation of normal 2 passing through C $(0,1)$ is
$\begin{align}
& y-1=1(x-0) \\
& y-1=x \\
\end{align}$
$x-y+1=0$ ..(5)
Now we have the equation of both tangent and normal to find the intersection point between tangent and normal respectively.
Solving equation (2) & (4) we get the point of intersection of tangent. Adding (2) & (4), we get $x=2$. Now putting the value of x in equation (2), we get $y=-1$. Let us name that point D $\left( 2,-1 \right)$.
Now solving equation (3) & (5) we get the intersection point of normal. Adding (3) & (5), we get $x=2$. And putting the value of y in equation (3), we get, $y=3$. Let us name that point E $\left( 2,3 \right)$.
Quadrilateral forms having coordinates B $\left( 4,1 \right)$, C $(0,1)$, D $\left( 2,-1 \right)$and E $\left( 2,3 \right)$. Observing the diagram on graph carefully, we know that tangent and normal are perpendicular to each other. So CD is perpendicular to the EC and BD is perpendicular to the EB. We know the distance formula, distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right)$ is equal to$\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$.
So CD $=2\sqrt{2}$
EC $=2\sqrt{2}$
BD $=2\sqrt{2}$
EB $=2\sqrt{2}$
ED $=4$
BC $=4$
So all sides are equal to quadrilateral and each angle makes ${{90}^{\circ }}$ and diagonal are also equal, so the resulting quadrilateral will be a square.
Area of square (quadrilateral) $=sid{{e}^{2}}$
Area of square (quadrilateral) $={{\left( 2\sqrt{2} \right)}^{2}}$ $=8$ sq. units.
Note: We can use an alternate method to determine the area of quadrilateral by dividing quadrilateral into two triangles and area of triangle can be found using coordinate geometry formula when the vertices of triangle points are given such that, for example when three vertices points are given such as $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right)$, then area is equal to $\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)$. Using that formula we can find the area of two triangles and then the area of the quadrilateral is the sum of the area of those two triangles.
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