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see attachment and explain it

 see attachment and explain it

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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

Curve 1:
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1
Differentiate
\frac{2x}{a^{2}} + \frac{2y}{b^{2}}.\frac{dy}{dx} = 0
Slope of tangent:
\frac{dy}{dx} = \frac{-b^{2}x}{a^{2}y}…......(1)
Curve 2:
\frac{x^{2}}{l^{2}} - \frac{y^{2}}{m^{2}} = 1
Differentiate
\frac{2x}{l^{2}} - \frac{2y}{m^{2}}.\frac{dy}{dx} = 0
Slope of tangent:
\frac{dy}{dx} = \frac{m^{2}x}{l^{2}y}...........(2)
Since curve intersect orthogonally, we have
(1)\times (2) = -1
\frac{b^{2}m^{2}x^{2}}{a^{2}l^{2}y^{2}} = 1
Since curve are intersecting, to find intersection
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{x^{2}}{l^{2}} - \frac{y^{2}}{m^{2}}
x^{2}(\frac{1}{a^{2}}-\frac{1}{l^{2}}) = y^{2}(-\frac{1}{b^{2}}-\frac{1}{m^{2}})
\frac{x^{2}}{y^{2}}(\frac{1}{a^{2}}-\frac{1}{l^{2}}) = (-\frac{1}{b^{2}}-\frac{1}{m^{2}})
\frac{a^{2}l^{2}}{b^{2}m^{2}}(\frac{1}{a^{2}}-\frac{1}{l^{2}}) = (-\frac{1}{b^{2}}-\frac{1}{m^{2}})
l^{2}-a^{2} = -m^{2}-b^{2}
a^{2} -b^{2} = l^{2} + m^{2}
Option © is correct.

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