The normal chord of the parabola ${{y}^{2}}=4ax$ at the point $\left( {{x}_{1}},{{x}_{1}} \right)$ subtends a right angle at the:
(a) focus.
(b) vertex.
(c) end of the latus rectum.
(d) none of these.
Answer
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Hint: We start solving the problem by drawing parabola and all the required points on it. We then find the point $\left( {{x}_{1}},{{x}_{1}} \right)$ using the parametric form of point on parabola. We then find the equation of the normal using the slope form of it. We then find the other point that the normal intersects the parabola. We then take the slope of the lines joining two intersection points of normal with parabola and verify whether the product of slopes is $-1$ to get the required result.
Complete step by step answer:
According to the parabola, we have a parabola ${{y}^{2}}=4ax$ and we need to find the point at which the normal chord to the parabola at the point $\left( {{x}_{1}},{{x}_{1}} \right)$ subtends right angle.
Let us draw the given information showing the parabola, given point, normal chord, focus, vertex, and ends of the latus rectum to get a better view.
We know that the vertex, focus and the ends of the latus rectum is $A\left( 0,0 \right)$, $B\left( a,0 \right)$, $\left( a,2a \right)$ and $\left( a,-2a \right)$. We also know that the point on the parabola is of the form $\left( a{{t}^{2}},2at \right)$. Let us find the point $P\left( {{x}_{1}},{{x}_{1}} \right)$ using this form.
So, we get $a{{t}^{2}}=2at$.
$\Rightarrow {{t}^{2}}=2t$.
$\Rightarrow {{t}^{2}}-2t=0$.
$\Rightarrow t\left( t-2 \right)=0$.
$\Rightarrow t=0$ or $t-2=0$.
$\Rightarrow t=2$.
We neglect $t=0$ as it gives the point as vertex at which we won’t get a normal chord.
So, the point $\left( {{x}_{1}},{{x}_{1}} \right)$ is $\left( a{{\left( 2 \right)}^{2}},2a\left( 2 \right) \right)=\left( 4a,4a \right)$.
We know that the slope form of the equation of the normal chord to the parabola ${{y}^{2}}=4ax$ is $y=mx-2am-a{{m}^{3}}$. Let us substitute the point $\left( 4a,4a \right)$ in this equation.
So, we get $4a=m\left( 4a \right)-2am-a{{m}^{3}}$.
$\Rightarrow 4a=4am-2am-a{{m}^{3}}$.
$\Rightarrow -a{{m}^{3}}+2am-4a=0$.
$\Rightarrow {{m}^{3}}-2m+4=0$.
$\Rightarrow {{m}^{3}}+2{{m}^{2}}-2{{m}^{2}}-4m+2m+4=0$.
$\Rightarrow {{m}^{2}}\left( m+2 \right)-2m\left( m+2 \right)+2\left( m+2 \right)=0$.
$\Rightarrow \left( m+2 \right)\left( {{m}^{2}}-2m+2 \right)=0$.
We can see that the discriminant of ${{m}^{2}}-2m+2=0$ is ${{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 2 \right)=-4<0$, which gives us no solution.
So, the value of m is $-2$ and the equation of the normal chord is $y=-2x-2a\left( -2 \right)-a{{\left( -2 \right)}^{3}}$.
$\Rightarrow y=-2x+4a+8a$.
$\Rightarrow 2x+y-12a=0$. Let us find the other point that this normal chord intersects the parabola ${{y}^{2}}=4ax$.
We substitute $y=12a-2x$ in ${{y}^{2}}=4ax$.
$\Rightarrow {{\left( 12a-2x \right)}^{2}}=4ax$.
$\Rightarrow 144{{a}^{2}}+4{{x}^{2}}-48ax=4ax$.
$\Rightarrow 4{{x}^{2}}-52ax+144{{a}^{2}}=0$.
$\Rightarrow {{x}^{2}}-13ax+36{{a}^{2}}=0$.
$\Rightarrow {{x}^{2}}-9ax-4ax+36{{a}^{2}}=0$.
$\Rightarrow x\left( x-9a \right)-4a\left( x-9a \right)=0$.
$\Rightarrow \left( x-4a \right)\left( x-9a \right)=0$.
$\Rightarrow x-4a=0$ or $x-9a=0$.
$\Rightarrow x=4a$ or $x=9a$.
Since $x=4a$ is the coordinate of the point where we find the equation of the normal chord. The x-coordinate of the other point is $x=9a$.
So, the y-coordinate of the other point is $y=12a-2\left( 9a \right)$.
$\Rightarrow y=12a-18a$.
$\Rightarrow y=-6a$.
So, the other point that the normal chord intersects the parabola is $Q\left( 9a,-6a \right)$.
Let us find the slope of the line joining vertex $A\left( 0,0 \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{1}}$.
$\Rightarrow {{m}_{1}}=\dfrac{4a-0}{4a-0}$.
$\Rightarrow {{m}_{1}}=\dfrac{4a}{4a}$.
$\Rightarrow {{m}_{1}}=1$ -(1).
Let us find the slope of the line joining vertex $A\left( 0,0 \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{2}}$.
$\Rightarrow {{m}_{2}}=\dfrac{-6a-0}{9a-0}$.
$\Rightarrow {{m}_{2}}=\dfrac{-6a}{9a}$.
$\Rightarrow {{m}_{2}}=\dfrac{-2}{3}$ -(2).
We know that the product of slopes of the two perpendiculars (non-vertical lines) is –1.
Here we get ${{m}_{1}}.{{m}_{2}}=1.\dfrac{-2}{3}=\dfrac{-2}{3}$, which is not equal to –1. This means that the normal chord doesn’t subtend right angle at vertex.
Now, let us find the slope of the line joining focus $B\left( a,0 \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{3}}$.
$\Rightarrow {{m}_{3}}=\dfrac{4a-0}{4a-a}$.
$\Rightarrow {{m}_{3}}=\dfrac{4a}{3a}$.
$\Rightarrow {{m}_{3}}=\dfrac{4}{3}$ -(3).
Let us find the slope of the line joining focus $B\left( a,0 \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{4}}$.
$\Rightarrow {{m}_{4}}=\dfrac{-6a-0}{9a-a}$.
$\Rightarrow {{m}_{4}}=\dfrac{-6a}{8a}$.
$\Rightarrow {{m}_{4}}=\dfrac{-3}{4}$ -(4).
We know that the product of slopes of the two perpendicular (non-vertical lines) is –1.
Here we get ${{m}_{3}}.{{m}_{4}}=\dfrac{4}{3}.\dfrac{-3}{4}=-1$, which is not equal to –1. This means that the normal chord subtends right angle at focus.
Now, let us find the slope of the line joining one of the ends of latus rectum $\left( a,2a \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{5}}$.
$\Rightarrow {{m}_{5}}=\dfrac{4a-2a}{4a-a}$.
$\Rightarrow {{m}_{5}}=\dfrac{2a}{3a}$.
$\Rightarrow {{m}_{5}}=\dfrac{2}{3}$ -(5).
Let us find the slope of the line joining one of the ends of latus rectum $\left( a,2a \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{6}}$.
$\Rightarrow {{m}_{6}}=\dfrac{-6a-2a}{9a-a}$.
$\Rightarrow {{m}_{6}}=\dfrac{-8a}{8a}$.
$\Rightarrow {{m}_{6}}=-1$ -(6).
We know that the product of slopes of the two perpendiculars (non-vertical lines) is –1.
Here we get ${{m}_{5}}.{{m}_{6}}=\dfrac{2}{3}.-1=\dfrac{-2}{3}$, which is not equal to –1. This means that the normal chord doesn’t subtends right angle at this end of latus rectum.
Now, let us find the slope of the line joining one of the ends of latus rectum $\left( a,-2a \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{7}}$.
$\Rightarrow {{m}_{7}}=\dfrac{4a+2a}{4a-a}$.
$\Rightarrow {{m}_{7}}=\dfrac{6a}{3a}$.
$\Rightarrow {{m}_{7}}=2$ -(7).
Let us find the slope of the line joining one of the ends of latus rectum $\left( a,-2a \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{8}}$.
$\Rightarrow {{m}_{8}}=\dfrac{-6a+2a}{9a-a}$.
$\Rightarrow {{m}_{8}}=\dfrac{-4a}{8a}$.
$\Rightarrow {{m}_{8}}=\dfrac{-1}{2}$ -(8).
We know that the product of slopes of the two perpendiculars (non-vertical lines) is –1.
Here we get ${{m}_{7}}.{{m}_{8}}=2.\dfrac{-1}{2}=-1$, which is equal to –1. This means that the normal chord subtends right angle at this end of the latus rectum.
So, we have found that the normal chord subtends a right angle at focus and at one of the ends of the latus rectum.
∴ The correct options for the given problem are (a) and (c).
Note:
We can also find the equation of the normal by finding the slope of the tangent using the differentiation and using the fact that normal is perpendicular to the tangent. We should check at both ends of the latus rectum in order to get the required answer. We can also find the area of the triangle formed between the intersection points of normal and any point given in the options.
Complete step by step answer:
According to the parabola, we have a parabola ${{y}^{2}}=4ax$ and we need to find the point at which the normal chord to the parabola at the point $\left( {{x}_{1}},{{x}_{1}} \right)$ subtends right angle.
Let us draw the given information showing the parabola, given point, normal chord, focus, vertex, and ends of the latus rectum to get a better view.
We know that the vertex, focus and the ends of the latus rectum is $A\left( 0,0 \right)$, $B\left( a,0 \right)$, $\left( a,2a \right)$ and $\left( a,-2a \right)$. We also know that the point on the parabola is of the form $\left( a{{t}^{2}},2at \right)$. Let us find the point $P\left( {{x}_{1}},{{x}_{1}} \right)$ using this form.
So, we get $a{{t}^{2}}=2at$.
$\Rightarrow {{t}^{2}}=2t$.
$\Rightarrow {{t}^{2}}-2t=0$.
$\Rightarrow t\left( t-2 \right)=0$.
$\Rightarrow t=0$ or $t-2=0$.
$\Rightarrow t=2$.
We neglect $t=0$ as it gives the point as vertex at which we won’t get a normal chord.
So, the point $\left( {{x}_{1}},{{x}_{1}} \right)$ is $\left( a{{\left( 2 \right)}^{2}},2a\left( 2 \right) \right)=\left( 4a,4a \right)$.
We know that the slope form of the equation of the normal chord to the parabola ${{y}^{2}}=4ax$ is $y=mx-2am-a{{m}^{3}}$. Let us substitute the point $\left( 4a,4a \right)$ in this equation.
So, we get $4a=m\left( 4a \right)-2am-a{{m}^{3}}$.
$\Rightarrow 4a=4am-2am-a{{m}^{3}}$.
$\Rightarrow -a{{m}^{3}}+2am-4a=0$.
$\Rightarrow {{m}^{3}}-2m+4=0$.
$\Rightarrow {{m}^{3}}+2{{m}^{2}}-2{{m}^{2}}-4m+2m+4=0$.
$\Rightarrow {{m}^{2}}\left( m+2 \right)-2m\left( m+2 \right)+2\left( m+2 \right)=0$.
$\Rightarrow \left( m+2 \right)\left( {{m}^{2}}-2m+2 \right)=0$.
We can see that the discriminant of ${{m}^{2}}-2m+2=0$ is ${{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 2 \right)=-4<0$, which gives us no solution.
So, the value of m is $-2$ and the equation of the normal chord is $y=-2x-2a\left( -2 \right)-a{{\left( -2 \right)}^{3}}$.
$\Rightarrow y=-2x+4a+8a$.
$\Rightarrow 2x+y-12a=0$. Let us find the other point that this normal chord intersects the parabola ${{y}^{2}}=4ax$.
We substitute $y=12a-2x$ in ${{y}^{2}}=4ax$.
$\Rightarrow {{\left( 12a-2x \right)}^{2}}=4ax$.
$\Rightarrow 144{{a}^{2}}+4{{x}^{2}}-48ax=4ax$.
$\Rightarrow 4{{x}^{2}}-52ax+144{{a}^{2}}=0$.
$\Rightarrow {{x}^{2}}-13ax+36{{a}^{2}}=0$.
$\Rightarrow {{x}^{2}}-9ax-4ax+36{{a}^{2}}=0$.
$\Rightarrow x\left( x-9a \right)-4a\left( x-9a \right)=0$.
$\Rightarrow \left( x-4a \right)\left( x-9a \right)=0$.
$\Rightarrow x-4a=0$ or $x-9a=0$.
$\Rightarrow x=4a$ or $x=9a$.
Since $x=4a$ is the coordinate of the point where we find the equation of the normal chord. The x-coordinate of the other point is $x=9a$.
So, the y-coordinate of the other point is $y=12a-2\left( 9a \right)$.
$\Rightarrow y=12a-18a$.
$\Rightarrow y=-6a$.
So, the other point that the normal chord intersects the parabola is $Q\left( 9a,-6a \right)$.
Let us find the slope of the line joining vertex $A\left( 0,0 \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{1}}$.
$\Rightarrow {{m}_{1}}=\dfrac{4a-0}{4a-0}$.
$\Rightarrow {{m}_{1}}=\dfrac{4a}{4a}$.
$\Rightarrow {{m}_{1}}=1$ -(1).
Let us find the slope of the line joining vertex $A\left( 0,0 \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{2}}$.
$\Rightarrow {{m}_{2}}=\dfrac{-6a-0}{9a-0}$.
$\Rightarrow {{m}_{2}}=\dfrac{-6a}{9a}$.
$\Rightarrow {{m}_{2}}=\dfrac{-2}{3}$ -(2).
We know that the product of slopes of the two perpendiculars (non-vertical lines) is –1.
Here we get ${{m}_{1}}.{{m}_{2}}=1.\dfrac{-2}{3}=\dfrac{-2}{3}$, which is not equal to –1. This means that the normal chord doesn’t subtend right angle at vertex.
Now, let us find the slope of the line joining focus $B\left( a,0 \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{3}}$.
$\Rightarrow {{m}_{3}}=\dfrac{4a-0}{4a-a}$.
$\Rightarrow {{m}_{3}}=\dfrac{4a}{3a}$.
$\Rightarrow {{m}_{3}}=\dfrac{4}{3}$ -(3).
Let us find the slope of the line joining focus $B\left( a,0 \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{4}}$.
$\Rightarrow {{m}_{4}}=\dfrac{-6a-0}{9a-a}$.
$\Rightarrow {{m}_{4}}=\dfrac{-6a}{8a}$.
$\Rightarrow {{m}_{4}}=\dfrac{-3}{4}$ -(4).
We know that the product of slopes of the two perpendicular (non-vertical lines) is –1.
Here we get ${{m}_{3}}.{{m}_{4}}=\dfrac{4}{3}.\dfrac{-3}{4}=-1$, which is not equal to –1. This means that the normal chord subtends right angle at focus.
Now, let us find the slope of the line joining one of the ends of latus rectum $\left( a,2a \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{5}}$.
$\Rightarrow {{m}_{5}}=\dfrac{4a-2a}{4a-a}$.
$\Rightarrow {{m}_{5}}=\dfrac{2a}{3a}$.
$\Rightarrow {{m}_{5}}=\dfrac{2}{3}$ -(5).
Let us find the slope of the line joining one of the ends of latus rectum $\left( a,2a \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{6}}$.
$\Rightarrow {{m}_{6}}=\dfrac{-6a-2a}{9a-a}$.
$\Rightarrow {{m}_{6}}=\dfrac{-8a}{8a}$.
$\Rightarrow {{m}_{6}}=-1$ -(6).
We know that the product of slopes of the two perpendiculars (non-vertical lines) is –1.
Here we get ${{m}_{5}}.{{m}_{6}}=\dfrac{2}{3}.-1=\dfrac{-2}{3}$, which is not equal to –1. This means that the normal chord doesn’t subtends right angle at this end of latus rectum.
Now, let us find the slope of the line joining one of the ends of latus rectum $\left( a,-2a \right)$ and the point $P\left( 4a,4a \right)$. Let it be ${{m}_{7}}$.
$\Rightarrow {{m}_{7}}=\dfrac{4a+2a}{4a-a}$.
$\Rightarrow {{m}_{7}}=\dfrac{6a}{3a}$.
$\Rightarrow {{m}_{7}}=2$ -(7).
Let us find the slope of the line joining one of the ends of latus rectum $\left( a,-2a \right)$ and the point $Q\left( 9a,-6a \right)$. Let it be ${{m}_{8}}$.
$\Rightarrow {{m}_{8}}=\dfrac{-6a+2a}{9a-a}$.
$\Rightarrow {{m}_{8}}=\dfrac{-4a}{8a}$.
$\Rightarrow {{m}_{8}}=\dfrac{-1}{2}$ -(8).
We know that the product of slopes of the two perpendiculars (non-vertical lines) is –1.
Here we get ${{m}_{7}}.{{m}_{8}}=2.\dfrac{-1}{2}=-1$, which is equal to –1. This means that the normal chord subtends right angle at this end of the latus rectum.
So, we have found that the normal chord subtends a right angle at focus and at one of the ends of the latus rectum.
∴ The correct options for the given problem are (a) and (c).
Note:
We can also find the equation of the normal by finding the slope of the tangent using the differentiation and using the fact that normal is perpendicular to the tangent. We should check at both ends of the latus rectum in order to get the required answer. We can also find the area of the triangle formed between the intersection points of normal and any point given in the options.
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