# Two tangents of the parabola \[{{y}^{2}}=4ax\] make angles \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\] with the \[x\]-axis. The locus of their point of intersection if \[\dfrac{\cot {{\alpha }_{1}}}{\cot {{\alpha }_{2}}}=2\] is

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

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

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

d) None of these

Answer

Verified

382.8k+ 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.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

What is 1 divided by 0 class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

What is pollution? How many types of pollution? Define it

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

How many crores make 10 million class 7 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE