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Question: Show that the two parabolas \( {x^2} + 4a\left( {y - 2b - a} \right) = 0{\text{ and }}{y^2} = 4b\lef...

Show that the two parabolas x2+4a(y−2b−a)=0 and y2=4b(x−2a+b){x^2} + 4a\left( {y - 2b - a} \right) = 0{\text{ and }}{y^2} = 4b\left( {x - 2a + b} \right) intersect at right angles at a common end of the latus rectum of each.

Explanation

Solution

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 (x−x0)2=4a(y−y0){\left( {x - {x_0}} \right)^2} = 4a\left( {y - {y_0}} \right)

⇒(x−x0)2=4a(y−y0) (1)\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 (x0+2a,y0+a) \left( {{x_0} + 2a,{y_0} + a} \right){\text{ }}

and (x0−2a,y0+a)\left( {{x_0} - 2a,{y_0} + a} \right)

And, other standard equation of parabola can be (x−x0)2=4a(y−y0){\left( {x - {x_0}} \right)^2} = 4a\left( {y - {y_0}} \right)

⇒(y−y0)2=4a(x−x0) (2)\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 (x0+a,y0+2a) \left( {{x_0} + a,{y_0} + 2a} \right){\text{ }}

and (x0+a,y0−2a)\left( {{x_0} + a,{y_0} - 2a} \right)

Given Equation of parabola are,

⇒x2=−4a(y−2b−a) (3) \Rightarrow {x^2} = - 4a\left( {y - 2b - a} \right){\text{ (3)}}

⇒y2=4b(x−2a+b) (4) \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 (−2a,2b+a−a) and (2a,2b+a−a)\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 (−2a,2b) and Q(2a,2b)\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 (2a−b+b,2b) and (2a−b+b,−2b)\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 (2a,2b) and S(2a,−2b)\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 (2a,2b) and R(2a,2b) \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=−4adydx2x = - 4a\dfrac{{dy}}{{dx}}

⇒dydx=−x2a=−2a2a=m1=−1 (Slope of parabola at equation 3)\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, 2ydydx=4b2y\dfrac{{dy}}{{dx}} = 4b

⇒dydx=2by=2b2b=m2=1 (Slope of parabola at equation 4) \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 m1m2=−1{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 . If m1 and m2{\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 m1m2=−1{m_1}{m_2} = - 1.