Questions & Answers

Question

Answers

a) \[2{{y}^{2}}=9ax\]

b) \[4{{y}^{2}}=9ax\]

c) \[{{y}^{2}}=9ax\]

d) None of these

Answer

Verified

156.9k+ views

Hint: To find the locus of point of intersection of two tangents to the parabola, we will write the equation of tangents in slope form and then use the angle formula to compare the slopes between two tangents.

Complete step-by-step answer:

We have a parabola of the form \[{{y}^{2}}=4ax\]. We have two tangents to the parabola which make angles \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\] with the \[x\]-axis such that \[\dfrac{\cot {{\alpha }_{1}}}{\cot {{\alpha }_{2}}}=2\] at their point of intersection.

We know that the equation of tangent of parabola of the form \[{{y}^{2}}=4ax\] with slope \[m\] is \[y=mx+\dfrac{a}{m}\].

If a line makes an angle \[\alpha \] with \[x\]-axis, the slope of the line is \[\tan \alpha \].

So, the two tangents making angles \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\] with \[x\]-axis have slopes \[\tan {{\alpha }_{1}}\] and \[\tan {{\alpha }_{2}}\].

The equation of tangent with slope \[\tan {{\alpha }_{1}}\] is \[y=\tan {{\alpha }_{1}}x+ \dfrac{a}{\tan {{\alpha }_{1}}}\].

Rewriting the above equation by dividing by \[\tan {{\alpha }_{1}}\], we get \[\dfrac{1}{\tan {{\alpha }_{1}}}\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=x\]. \[\left( 1 \right)\]

The equation of tangent with slope \[\tan {{\alpha }_{2}}\] is \[y=\tan {{\alpha }_{2}}x+ \dfrac{a}{\tan {{\alpha }_{2}}}\].

Rewriting the above equation by dividing by \[\tan {{\alpha }_{2}}\], we get \[\dfrac{1}{\tan {{\alpha }_{2}}}\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)=x\]. \[\left( 2 \right)\]

To find the locus of their point of intersection, we will solve the two equations in terms of \[x\] and \[y\].

Eliminating \[x\] from the two equations and equating them, we get \[\dfrac{1}{\tan {{\alpha }_{1}}}\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=\dfrac{1}{\tan {{\alpha }_{2}}}\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)\] \[\left( 3 \right)\]

We know, \[\dfrac{\cot {{\alpha }_{1}}}{\cot {{\alpha }_{2}}}=2\Rightarrow \dfrac{\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}=2\]

Multiplying equation \[\left( 3 \right)\] by \[\tan {{\alpha }_{2}}\], we get \[\dfrac{\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)\]

\[\Rightarrow 2\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)\]

On further solving the equation by taking LCM on both sides multiplying by \[\tan {{\alpha }_{2}}\], we get \[2\dfrac{\tan {{\alpha }_{2}}\left( y\tan {{\alpha }_{1}}-a \right)}{\tan {{\alpha }_{1}}}=y\tan {{\alpha }_{2}}-a\]

As we know \[\dfrac{\cot {{\alpha }_{1}}}{\cot {{\alpha }_{2}}}=2\Rightarrow \dfrac{\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}=2\], we get\[4\left( y\tan {{\alpha }_{1}}-a \right)=y\tan {{\alpha }_{2}}-a\]

\[\Rightarrow 4y\tan {{\alpha }_{1}}-4a=y\tan {{\alpha }_{2}}-a\]

Dividing the equation by \[\tan {{\alpha }_{1}}\], we get \[4y-\dfrac{4a}{\tan {{\alpha }_{1}}}=\dfrac{y\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}-\dfrac{a}{\tan {{\alpha }_{1}}}\]

\[\Rightarrow 4y-\dfrac{4a}{\tan {{\alpha }_{1}}}=2y-\dfrac{a}{\tan {{\alpha }_{1}}}\]

\[\Rightarrow 2y=\dfrac{3a}{\tan {{\alpha }_{1}}}\]

Rearranging the above equation, we get \[\tan {{\alpha }_{1}}=\dfrac{3a}{2y}\]

Substituting the value of \[\tan {{\alpha }_{1}}\] in equation \[\left( 1 \right)\], we get \[y=\dfrac{3a}{2y}x+\dfrac{a}{\dfrac{3a}{2y}}\]

\[\Rightarrow y=\dfrac{3ax}{2y}+\dfrac{2y}{3}\]

\[\Rightarrow \dfrac{y}{3}=\dfrac{3ax}{2y}\]

\[\begin{align}

& \Rightarrow 2{{y}^{2}}=9ax \\

& \\

\end{align}\]

We observe that this curve is the equation of a parabola. However, it is not necessary that we will always get a parabola. If we change the angle between two tangents, we will get a different curve. If the two tangents are perpendicular to each other, we will get locus as a straight line.

Hence, the correct answer is a) \[2{{y}^{2}}=9ax\].

Note: We can also solve this question by taking the point of intersection of tangents as any general point and then writing the equation of tangents at the general point.

Complete step-by-step answer:

We have a parabola of the form \[{{y}^{2}}=4ax\]. We have two tangents to the parabola which make angles \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\] with the \[x\]-axis such that \[\dfrac{\cot {{\alpha }_{1}}}{\cot {{\alpha }_{2}}}=2\] at their point of intersection.

We know that the equation of tangent of parabola of the form \[{{y}^{2}}=4ax\] with slope \[m\] is \[y=mx+\dfrac{a}{m}\].

If a line makes an angle \[\alpha \] with \[x\]-axis, the slope of the line is \[\tan \alpha \].

So, the two tangents making angles \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\] with \[x\]-axis have slopes \[\tan {{\alpha }_{1}}\] and \[\tan {{\alpha }_{2}}\].

The equation of tangent with slope \[\tan {{\alpha }_{1}}\] is \[y=\tan {{\alpha }_{1}}x+ \dfrac{a}{\tan {{\alpha }_{1}}}\].

Rewriting the above equation by dividing by \[\tan {{\alpha }_{1}}\], we get \[\dfrac{1}{\tan {{\alpha }_{1}}}\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=x\]. \[\left( 1 \right)\]

The equation of tangent with slope \[\tan {{\alpha }_{2}}\] is \[y=\tan {{\alpha }_{2}}x+ \dfrac{a}{\tan {{\alpha }_{2}}}\].

Rewriting the above equation by dividing by \[\tan {{\alpha }_{2}}\], we get \[\dfrac{1}{\tan {{\alpha }_{2}}}\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)=x\]. \[\left( 2 \right)\]

To find the locus of their point of intersection, we will solve the two equations in terms of \[x\] and \[y\].

Eliminating \[x\] from the two equations and equating them, we get \[\dfrac{1}{\tan {{\alpha }_{1}}}\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=\dfrac{1}{\tan {{\alpha }_{2}}}\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)\] \[\left( 3 \right)\]

We know, \[\dfrac{\cot {{\alpha }_{1}}}{\cot {{\alpha }_{2}}}=2\Rightarrow \dfrac{\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}=2\]

Multiplying equation \[\left( 3 \right)\] by \[\tan {{\alpha }_{2}}\], we get \[\dfrac{\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)\]

\[\Rightarrow 2\left( y-\dfrac{a}{\tan {{\alpha }_{1}}} \right)=\left( y-\dfrac{a}{\tan {{\alpha }_{2}}} \right)\]

On further solving the equation by taking LCM on both sides multiplying by \[\tan {{\alpha }_{2}}\], we get \[2\dfrac{\tan {{\alpha }_{2}}\left( y\tan {{\alpha }_{1}}-a \right)}{\tan {{\alpha }_{1}}}=y\tan {{\alpha }_{2}}-a\]

As we know \[\dfrac{\cot {{\alpha }_{1}}}{\cot {{\alpha }_{2}}}=2\Rightarrow \dfrac{\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}=2\], we get\[4\left( y\tan {{\alpha }_{1}}-a \right)=y\tan {{\alpha }_{2}}-a\]

\[\Rightarrow 4y\tan {{\alpha }_{1}}-4a=y\tan {{\alpha }_{2}}-a\]

Dividing the equation by \[\tan {{\alpha }_{1}}\], we get \[4y-\dfrac{4a}{\tan {{\alpha }_{1}}}=\dfrac{y\tan {{\alpha }_{2}}}{\tan {{\alpha }_{1}}}-\dfrac{a}{\tan {{\alpha }_{1}}}\]

\[\Rightarrow 4y-\dfrac{4a}{\tan {{\alpha }_{1}}}=2y-\dfrac{a}{\tan {{\alpha }_{1}}}\]

\[\Rightarrow 2y=\dfrac{3a}{\tan {{\alpha }_{1}}}\]

Rearranging the above equation, we get \[\tan {{\alpha }_{1}}=\dfrac{3a}{2y}\]

Substituting the value of \[\tan {{\alpha }_{1}}\] in equation \[\left( 1 \right)\], we get \[y=\dfrac{3a}{2y}x+\dfrac{a}{\dfrac{3a}{2y}}\]

\[\Rightarrow y=\dfrac{3ax}{2y}+\dfrac{2y}{3}\]

\[\Rightarrow \dfrac{y}{3}=\dfrac{3ax}{2y}\]

\[\begin{align}

& \Rightarrow 2{{y}^{2}}=9ax \\

& \\

\end{align}\]

We observe that this curve is the equation of a parabola. However, it is not necessary that we will always get a parabola. If we change the angle between two tangents, we will get a different curve. If the two tangents are perpendicular to each other, we will get locus as a straight line.

Hence, the correct answer is a) \[2{{y}^{2}}=9ax\].

Note: We can also solve this question by taking the point of intersection of tangents as any general point and then writing the equation of tangents at the general point.