Question

Show the condition that the curves \[a{x^2} + b{y^2} = 1\] and \[{a^,}{x^2} + {b^,}{y^2} = 1\] should intersect orthogonally is \[\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{{{a^,}}} - \dfrac{1}{{{b^,}}}\] .

Answer 1

Hint: When any of the two curves are intersecting orthogonally , then the angle between them is \[{90^ \circ }\] . The condition for any two curves are orthogonal , is \[{m_1}{m_2} = - 1\] , where \[{m_1}\] and \[{m_2}\] are the slopes of the curves . The slope can be obtained by differentiating the curve .

Complete step-by-step answer:
Given : \[a{x^2} + b{y^2} = 1\] …….. (a) and \[{a^,}{x^2} + {b^,}{y^2} = 1\] …… (b)
Differentiating equation (a) with respect to \[x\] , we get the slope for curve A
\[2ax + 2b\dfrac{{dy}}{{dx}} = 0\] , on solving we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{ - ax}}{{by}}\]
Similarly , differentiating the equation (b) with respect to \[x\] , we get the slope of the curve B
\[2{a^,}x + 2{b^,}y\dfrac{{dy}}{{dx}} = 0\] , on simplifying we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{ - {a^,}x}}{{{b^,}y}}\]
Since , the curves intersect orthogonally therefore , the product of the slopes will be \[ - 1\] .
Hence ,
\[\left( {\dfrac{{ - ax}}{{by}}} \right)\left( {\dfrac{{ - {a^,}x}}{{{b^,}y}}} \right) = - 1\] , on solving we get
\[{x^2}a{a^,} = - b{b^,}{y^2}\] …..(i)
Now , subtracting equation (b) from (a) , we get
\[\left( {a - {a^,}} \right){x^2} + \left( {b - {b^,}} \right){y^2} = 0\] ……(ii)
Dividing equation (ii) by (i) , we get
\[\dfrac{{\left( {a - {a^,}} \right){x^2}}}{{{x^2}a{a^,}}} = \dfrac{{ - \left( {b - {b^,}} \right){y^2}}}{{ - b{b^,}{y^2}}}\] ,
\[\dfrac{{\left( {a - {a^,}} \right)}}{{a{a^,}}} = \dfrac{{\left( {b - {b^,}} \right)}}{{b{b^,}}}\] , on solving we get
\[\dfrac{1}{{{a^,}}} - \dfrac{1}{a} = \dfrac{1}{{{b^,}}} - \dfrac{1}{b}\] , on simplifying we get
\[\dfrac{1}{{{a^,}}} - \dfrac{1}{{{b^,}}} = \dfrac{1}{a} - \dfrac{1}{b}\] .
Hence , the condition for orthogonality is proved .

Note: These questions are the applications of differentiation , where slope is obtained using the derivative of the curve . Actually , it gives the slope of the tangent to the curve . The tangent is a straight line which just touches the curve at a given point. The normal is a straight line which is perpendicular to the tangent . Remember the condition when the curves are orthogonal and parallel . For parallel the product of the will be equal to zero .