# The locus of point of intersection of tangents to the parabolas \[{y^2} = 4\left( {x + 1} \right)\]and \[{y^2} = 8\left( {x + 2} \right)\] which are perpendicular to each other is

(a)$x + 7 = 0$ (b) $x - y = 4$

(c)$x + 3 = 0$ (d)$y - x = 12$

Last updated date: 27th Mar 2023

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Hint: Find the general equation of tangent at any point on the parabola using the slope-point form$\left( {y - y_1} \right) = m\left( {x - x_1} \right)$. Compare that with the given parabola equations to find the point of tangency for both the tangents. Use the condition for the two tangents to be perpendicular to find the locus of the point of intersection.

The given equations are

\[{y^2} = 4\left( {x + 1} \right)\] …(1)

\[{y^2} = 8\left( {x + 2} \right)\] …(2)

The general form of a parabola with its vertex at the origin and its axis parallel to the x-axis is ${y^2} = 4ax$ …(3)

Point of tangency is given by $\left( {a{t^2},2at} \right)$

We get the slope by differentiating with respect to$x$.

So, let us differentiate equation (3) to find its slope and use the slope point form to find the equation of tangent at$\left( {a{t^2},2at} \right)$.

$

{y^2} = 4ax \\

2y\dfrac{{dy}}{{dx}} = 4a \\

\dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y} \\

$

At point$\left( {a{t^2},2at} \right)$,$y = 2at$. Substitute to get slope$m$from the differentiated value.

$

\dfrac{{dy}}{{dx}} = \dfrac{{2a}}{{2at}} \\

m = \dfrac{1}{t} \\

$

Slope Point form to find the equation of a line passing through a point $\left( {x_1,y_1} \right)$ with a slope$m$is written as $\left( {y - y_1} \right) = m\left( {x - x_1} \right)$

So, general equation of a tangent line at point $\left( {a{t^2},2at} \right)$and with slope $m = \dfrac{1}{t}$ is given by

$

y - 2at = \dfrac{1}{t}\left( {x - a{t^2}} \right) \\

ty - 2a{t^2} = x - a{t^2} \\

$

$ty = x + a{t^2}$ …(5)

From equation (1), comparing with the general form of parabola in equation (3), we get

$a = 4,x = x + 1$ …(6)

So the point of tangency for (1) is $\left( {t{1^2},2t_1} \right)$ …(7)

Similarly from equation (2),

$a = 8,x = x + 2$ …(8)

So the point of tangency for (2) is $\left( {2t{2^2},4t_2} \right)$ …(9)

For finding tangent equation for parabola in equation (1) substitute (6) and (7) in (5)

$t_1y = \left( {x + 1} \right) + t{1^2}$

$y = \dfrac{x}{{t_1}} + \left( {t_1 + \dfrac{1}{{t_2}}} \right)$ …(10)

For finding tangent equation for parabola in equation (2) substitute (8) and (9) in (5)

$t_2y = \left( {x + 2} \right) + 2t{2^2}$

$y = \dfrac{x}{{t_2}} + 2\left( {t_2 + \dfrac{1}{{t_2}}} \right)$ …(11)

The tangents (10) and (11) are perpendicular (given)

So, the product of their slope is -1.

If $m_1,m_2$are their respective slope,

$m_1m_2 = - 1$

$

m_1 = \dfrac{1}{{t_1}},m_2 = \dfrac{1}{{t_2}} \\

\dfrac{1}{{t_1}}\dfrac{1}{{t_2}} = - 1 \\

t_1t_2 = - 1 \\

$

Equating (10) and (11) to find their point of intersection and substituting \[t_1t_2 = - 1\]

$

\dfrac{x}{{t_1}} + \left( {t_1 + \dfrac{1}{{t_1}}} \right) = \dfrac{x}{{t_2}} + 2\left( {t_2 + \dfrac{1}{{t_2}}} \right) \\

x\left( {\dfrac{1}{{t_1}} - \dfrac{1}{{t_2}}} \right) = 2t_2 + \dfrac{2}{{t_2}} - t_1 - \dfrac{1}{{t_1}} \\

x\left( {\dfrac{{t_2 - t_1}}{{t_1t_2}}} \right) = \dfrac{{2t{2^2}t_1 + 2t_1 - t{1^2}t_2 - t_2}}{{t_1t_2}} \\

x\left( {t_2 - t_1} \right) = - 2t_2 + 2t_1 + t_1 - t_2 \\

x\left( {t_2 - t_1} \right) = - 3\left( {t_2 - t_1} \right) \\

x = - 3 \\

x + 3 = 0 \\

$

Hence the locus of the point of intersection of the tangents is \[x + 3 = 0\]

Hence, the correct answer is option (c).

Note: We are finding the equation of tangents for both the given parabolas and finding their point of intersection with the condition that they are perpendicular, to find the locus of that point of intersection. We can also do this without finding the general equation of tangent and then substituting. Instead, we can directly find the equation of tangents for each parabola and then equate it to find the point of intersection.

The given equations are

\[{y^2} = 4\left( {x + 1} \right)\] …(1)

\[{y^2} = 8\left( {x + 2} \right)\] …(2)

The general form of a parabola with its vertex at the origin and its axis parallel to the x-axis is ${y^2} = 4ax$ …(3)

Point of tangency is given by $\left( {a{t^2},2at} \right)$

We get the slope by differentiating with respect to$x$.

So, let us differentiate equation (3) to find its slope and use the slope point form to find the equation of tangent at$\left( {a{t^2},2at} \right)$.

$

{y^2} = 4ax \\

2y\dfrac{{dy}}{{dx}} = 4a \\

\dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y} \\

$

At point$\left( {a{t^2},2at} \right)$,$y = 2at$. Substitute to get slope$m$from the differentiated value.

$

\dfrac{{dy}}{{dx}} = \dfrac{{2a}}{{2at}} \\

m = \dfrac{1}{t} \\

$

Slope Point form to find the equation of a line passing through a point $\left( {x_1,y_1} \right)$ with a slope$m$is written as $\left( {y - y_1} \right) = m\left( {x - x_1} \right)$

So, general equation of a tangent line at point $\left( {a{t^2},2at} \right)$and with slope $m = \dfrac{1}{t}$ is given by

$

y - 2at = \dfrac{1}{t}\left( {x - a{t^2}} \right) \\

ty - 2a{t^2} = x - a{t^2} \\

$

$ty = x + a{t^2}$ …(5)

From equation (1), comparing with the general form of parabola in equation (3), we get

$a = 4,x = x + 1$ …(6)

So the point of tangency for (1) is $\left( {t{1^2},2t_1} \right)$ …(7)

Similarly from equation (2),

$a = 8,x = x + 2$ …(8)

So the point of tangency for (2) is $\left( {2t{2^2},4t_2} \right)$ …(9)

For finding tangent equation for parabola in equation (1) substitute (6) and (7) in (5)

$t_1y = \left( {x + 1} \right) + t{1^2}$

$y = \dfrac{x}{{t_1}} + \left( {t_1 + \dfrac{1}{{t_2}}} \right)$ …(10)

For finding tangent equation for parabola in equation (2) substitute (8) and (9) in (5)

$t_2y = \left( {x + 2} \right) + 2t{2^2}$

$y = \dfrac{x}{{t_2}} + 2\left( {t_2 + \dfrac{1}{{t_2}}} \right)$ …(11)

The tangents (10) and (11) are perpendicular (given)

So, the product of their slope is -1.

If $m_1,m_2$are their respective slope,

$m_1m_2 = - 1$

$

m_1 = \dfrac{1}{{t_1}},m_2 = \dfrac{1}{{t_2}} \\

\dfrac{1}{{t_1}}\dfrac{1}{{t_2}} = - 1 \\

t_1t_2 = - 1 \\

$

Equating (10) and (11) to find their point of intersection and substituting \[t_1t_2 = - 1\]

$

\dfrac{x}{{t_1}} + \left( {t_1 + \dfrac{1}{{t_1}}} \right) = \dfrac{x}{{t_2}} + 2\left( {t_2 + \dfrac{1}{{t_2}}} \right) \\

x\left( {\dfrac{1}{{t_1}} - \dfrac{1}{{t_2}}} \right) = 2t_2 + \dfrac{2}{{t_2}} - t_1 - \dfrac{1}{{t_1}} \\

x\left( {\dfrac{{t_2 - t_1}}{{t_1t_2}}} \right) = \dfrac{{2t{2^2}t_1 + 2t_1 - t{1^2}t_2 - t_2}}{{t_1t_2}} \\

x\left( {t_2 - t_1} \right) = - 2t_2 + 2t_1 + t_1 - t_2 \\

x\left( {t_2 - t_1} \right) = - 3\left( {t_2 - t_1} \right) \\

x = - 3 \\

x + 3 = 0 \\

$

Hence the locus of the point of intersection of the tangents is \[x + 3 = 0\]

Hence, the correct answer is option (c).

Note: We are finding the equation of tangents for both the given parabolas and finding their point of intersection with the condition that they are perpendicular, to find the locus of that point of intersection. We can also do this without finding the general equation of tangent and then substituting. Instead, we can directly find the equation of tangents for each parabola and then equate it to find the point of intersection.

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