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# Show that the two parabolas ${x^2} + 4a\left( {y - 2b - a} \right) = 0$ and ${y^2} = 4b\left( {x - 2a + b} \right)$ intersect at right angles at a common end of the latus rectum of each.

Last updated date: 20th Jun 2024
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Hint: Draw the given parabolas and find the intersection points and use properties of parabola to prove the given condition.

Complete step-by-step answer:

As, we know that standard equation of parabola is ${\left( {x - {x_0}} \right)^2} = 4a\left( {y - {y_0}} \right)$

$\Rightarrow {\left( {x - {x_0}} \right)^2} = 4a\left( {y - {y_0}} \right){\text{ (1)}}$

As, we know coordinates of end points of latus rectum of equation 1 will be $\left( {{x_0} + 2a,{y_0} + a} \right){\text{ }}$

and  $\left( {{x_0} - 2a,{y_0} + a} \right)$

And, other standard equation of parabola can be ${\left( {x - {x_0}} \right)^2} = 4a\left( {y - {y_0}} \right)$

$\Rightarrow {\left( {y - {y_0}} \right)^2} = 4a\left( {x - {x_0}} \right){\text{ (2)}}$

$\Rightarrow$ As, we know coordinates of end points of latus rectum of equation 2 will be $\left( {{x_0} + a,{y_0} + 2a} \right){\text{ }}$

and $\left( {{x_0} + a,{y_0} - 2a} \right)$

Given Equation of parabola are,

$\Rightarrow {x^2} = - 4a\left( {y - 2b - a} \right){\text{ (3)}}$

$\Rightarrow {y^2} = 4b\left( {x - 2a + b} \right){\text{ (4) }}$

On comparing equation 3 with equation 1 we get,

Coordinates of endpoints of latus rectum of equation 3 will be $\left( { - 2a,2b + a - a} \right){\text{ and }}\left( {2a,2b + a - a} \right)$

On solving coordinates of endpoints of latus rectum of equation 3 will be P $\left( { - 2a,2b} \right){\text{ and Q}}\left( {2a,2b} \right)$

On comparing equation 4 with equation 2 we get,

Coordinates of endpoints of latus rectum of equation 4 will be $\left( {2a - b + b,2b} \right){\text{ and }}\left( {2a - b + b, - 2b} \right)$

On solving coordinates of endpoints of latus rectum of equation 4 will be R $\left( {2a,2b} \right){\text{ and S}}\left( {2a, - 2b} \right)$

As we see, common end to the latus rectum of parabola at equation 3 and 4 is Q $\left( {2a,2b} \right){\text{ and R}}\left( {2a,2b} \right){\text{ }}$

For, finding the slope of parabola in equation 3 we have to differentiate equation 3 w.r.t x

Differentiating equation 3 we get, $2x = - 4a\dfrac{{dy}}{{dx}}$

$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{{2a}} = \dfrac{{ - 2a}}{{2a}} = {m_1} = - 1{\text{ }}\left( {{\text{Slope of parabola at equation 3}}} \right)$

For, finding the slope of parabola in equation 4 we have to differentiate equation 4 w.r.t x

Differentiating equation 4 we get, $2y\dfrac{{dy}}{{dx}} = 4b$

$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2b}}{y} = \dfrac{{2b}}{{2b}} = {m_2} = 1{\text{ }}\left( {{\text{Slope of parabola at equation 4}}} \right){\text{ }}$

As we can see that ${m_1}{m_2} = - 1$

Hence both the given parabolas intersect at right angles at a common end of the latus rectum.

Hence proved.

NOTE: - Whenever you come up with this type of problem the best way is to find the endpoints of the latus rectum and then find the angle between them ${\text{. If }}{m_1}{\text{ and }}{m_2}$ are the slope of two curves at at some point, then the both intersect at right angle only if ${m_1}{m_2} = - 1$.