If two tangents drawn from a point $P$ to the parabola ${y^2} = 4x$ are at right angles, then the locus of $P$ is:
A) $2x + 1 = 0$
B) $x = - 1$
C) $2x - 1 = 0$
D) $x = 1$
VerifiedHint: We can write the equation of tangent to a parabola as $y = mx + \dfrac{a}{m}$. Where $m$ is the slope and here we write parabola in the form ${y^2} = 4ax$. Use these identities to get the desired answer.
Complete step-by-step answer:
Here a parabola is given ${y^2} = 4x$. And it is said that there is a point $P$ from which tangents drawn to parabola make right angles with each other.
Let the coordinates of point $P$ be $\left( {h,k} \right)$. And let $P{T_1},P{T_2}$ be the two tangents such that it is given that both tangents are perpendicular to each other which means $\angle {T_1}P{T_2} = {90^ \circ }$.
We know if two lines with slope ${m_1},{m_2}$ are perpendicular, then ${m_1}{m_2} = - 1$.
Also we know that tangent to the parabola ${y^2} = 4ax$ is given as $y = mx + \dfrac{a}{m}$.
Here let $P{T_1}$ has slope ${m_1}$. And $a = 1$
So, equation of tangent $P{T_1}$
$y = {m_1}x + \dfrac{1}{{{m_1}}}$ (1)
And equation of $P{T_2}$ with slope ${m_2}$can be written as
$y = {m_2}x + \dfrac{1}{{{m_2}}}$ (2)
We know that ${m_1}{m_2} = - 1$
$ \Rightarrow {m_1} = - \dfrac{1}{{{m_2}}}$
Putting it in equation (1)
$y = - \dfrac{x}{{{m_2}}} - {m_2}$
And we know
$y = {m_2}x + \dfrac{1}{{{m_2}}}$
So,
${m_2}x + \dfrac{1}{{{m_2}}} = - \left( {\dfrac{x}{{{m_2}}} + {m_2}} \right)$
$\left( {{m_2} + \dfrac{1}{{{m_2}}}} \right)x = - \left( {{m_2} + \dfrac{1}{{{m_2}}}} \right)$
$x = - 1 $
So, option B is correct.
Note: If two lines are given, $y = {m_1}x + {c_1}{\text{ and }}y = {m_2}x + {c_2}$. Then angle between these lines can be given by
$\tan \theta = \dfrac{{|{m_1} - {m_2}|}}{{|1 + {m_1}{m_2}|}}$
If both are perpendicular, then $\tan {90^ \circ } = \infty $. So,
$1 + {m_1}{m_2} = 0$
${m_1}{m_2} = - 1 $
If both are parallel, then $\tan {0^ \circ } = 0$. So,
${m_1} - {m_2} = 0 $
${m_1} = {m_2} $
