Answer
457.5k+ views
Hint:To find the locus of point P, write the equation of chord of contact of tangents to the parabola ${{y}^{2}}=4ax$. Also, write the equation of tangent to the parabola ${{x}^{2}}=4by$. Compare the both equations of tangents and simplify them to calculate the locus of point P.
Complete step-by-step answer:
We have to find the locus of point P from which the chord of contact of tangents of the parabola \[{{y}^{2}}=4ax\] touches the parabola \[{{x}^{2}}=4by\]. We will write the equation of tangent to the parabola \[{{x}^{2}}=4by\] and equation of chord of contact of tangents to the parabola \[{{y}^{2}}=4ax\] from any general point P and then compare the two equation of lines.
We know the equation of tangent of parabola \[{{x}^{2}}=4by\] with slope m is \[x=my+\dfrac{b}{m}\].
Rewriting the equation of tangent, we get \[y=\dfrac{x}{m}-\dfrac{b}{{{m}^{2}}}.....\left( 1 \right)\].
We know that the equation of chord of contact of parabola \[{{y}^{2}}=4ax\] drawn from a point \[P(u,v)\] is of the form \[yv=2a(x+u)\].
Rearranging the terms, we have \[y=\dfrac{2ax}{v}+\dfrac{2au}{v}.....\left( 2 \right)\]
Equation (1) and (2) represent the same line.
Comparing the slope and intercept of both equations, we get \[\dfrac{1}{m}=\dfrac{2a}{v},\dfrac{-b}{{{m}^{2}}}=\dfrac{2au}{v}\].
To solve the above equations, multiply b to the square of the first equation and add to the second equation.
Thus, we have
\[\begin{align}
& \Rightarrow \dfrac{b}{{{m}^{2}}}+\dfrac{-b}{{{m}^{2}}}=\dfrac{4{{a}^{2}}}{{{v}^{2}}}+\dfrac{2au}{v} \\
& \Rightarrow 2au=\dfrac{-4{{a}^{2}}}{{{v}^{2}}}v \\
& \Rightarrow uv=-2a \\
\end{align}\]
Hence, the locus of \[P(u,v)\] from which chord of contact of tangents is drawn to the parabola is a hyperbola, which is option (d).
Note: It’s necessary to compare the slope and intercept of both the lines and simplify them to find the locus of point P. We can also solve this question by writing the equation of line in point slope form and then comparing it with another equation of line to find the locus.
Complete step-by-step answer:
![seo images](https://www.vedantu.com/question-sets/8add16d6-ae84-40ed-9488-9c510f912b1a4022062603899748143.png)
We have to find the locus of point P from which the chord of contact of tangents of the parabola \[{{y}^{2}}=4ax\] touches the parabola \[{{x}^{2}}=4by\]. We will write the equation of tangent to the parabola \[{{x}^{2}}=4by\] and equation of chord of contact of tangents to the parabola \[{{y}^{2}}=4ax\] from any general point P and then compare the two equation of lines.
We know the equation of tangent of parabola \[{{x}^{2}}=4by\] with slope m is \[x=my+\dfrac{b}{m}\].
Rewriting the equation of tangent, we get \[y=\dfrac{x}{m}-\dfrac{b}{{{m}^{2}}}.....\left( 1 \right)\].
We know that the equation of chord of contact of parabola \[{{y}^{2}}=4ax\] drawn from a point \[P(u,v)\] is of the form \[yv=2a(x+u)\].
Rearranging the terms, we have \[y=\dfrac{2ax}{v}+\dfrac{2au}{v}.....\left( 2 \right)\]
Equation (1) and (2) represent the same line.
Comparing the slope and intercept of both equations, we get \[\dfrac{1}{m}=\dfrac{2a}{v},\dfrac{-b}{{{m}^{2}}}=\dfrac{2au}{v}\].
To solve the above equations, multiply b to the square of the first equation and add to the second equation.
Thus, we have
\[\begin{align}
& \Rightarrow \dfrac{b}{{{m}^{2}}}+\dfrac{-b}{{{m}^{2}}}=\dfrac{4{{a}^{2}}}{{{v}^{2}}}+\dfrac{2au}{v} \\
& \Rightarrow 2au=\dfrac{-4{{a}^{2}}}{{{v}^{2}}}v \\
& \Rightarrow uv=-2a \\
\end{align}\]
Hence, the locus of \[P(u,v)\] from which chord of contact of tangents is drawn to the parabola is a hyperbola, which is option (d).
Note: It’s necessary to compare the slope and intercept of both the lines and simplify them to find the locus of point P. We can also solve this question by writing the equation of line in point slope form and then comparing it with another equation of line to find the locus.
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