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TANGENTS AND NORMALS show that the curves x^2 / (a^2 + ​​(lambda) 1 ) + y^2 / (b^2 + (lambda) 1 ) =1 and x^2 / (a^2 + (lambda) 2 ) + y^2 / (b^2 + (lambda) 2 ) = 1 intersect at right angles

TANGENTS AND NORMALS
 
show that the curves x^2 / (a^2 + ​​(lambda)1) + y^2 / (b^2 + (lambda)1) =1  and x^2 / (a^2 + (lambda)2) + y^2 / (b^2 + (lambda)2) = 1 intersect at right angles

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

\frac{x^2}{a^2+\lambda _{1}} + \frac{y^2}{b^2+\lambda _{1}} = 1
Slope of tangent:
\frac{2x}{a^2+\lambda _{1}} + \frac{2yy'_{1}}{b^2+\lambda _{1}} = 0
y'_{1} = \frac{-x(b^2+\lambda _{1})}{y(a^2+\lambda _{1})}.................(1)
\frac{x^2}{a^2 + \lambda _{2}} + \frac{y^2}{b^2+\lambda _{2}} = 1
Slope of tangent:
\frac{2x}{a^2+\lambda _{2}} + \frac{2yy'_{2}}{b^2+\lambda _{2}} = 0
y'_{2} = \frac{-x(b^2+\lambda _{2})}{y(a^2+\lambda _{2})}..............(2)
(1)\times (2) = -1
y_{1}'y_{2}' = \frac{x^2(b^2+\lambda _{1})(b^2+\lambda _{2})}{y^2(a^2+\lambda _{2})(a^2+\lambda _{2})}
Just manipulate the curve equations from here for x2/y2.
y'_{1}y'_{2} = -1

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