Find condition that the curves ax^2+by^2=1 and a1x^2+b1y^2=1intersect orthogonally

Find condition that the curves ax^2+by^2=1 and a1x^2+b1y^2=1intersect orthogonally
Find the condition that the curves intersect orthogonally
To find the condition for the curves \(ax^2 + by^2 = 1\) and \(a_1x^2 + b_1y^2 = 1\) to intersect orthogonally, we can use the fact that the product of the slopes of two curves at their point of intersection should be -1.
Chapter plane curve::
The general form of a curve is \(y = f(x)\), and its slope is given by \(f'(x)\). For the curves \(ax^2 + by^2 = 1\) and \(a_1x^2 + b_1y^2 = 1\), we need to find their derivatives with respect to \(x\), and then set the product of their slopes equal to -1.
This leads to the condition:
\[2ax + 2b\frac{dy}{dx} \cdot \frac{-1}{2b} + 2a_1x + 2b_1\frac{dy}{dx} \cdot \frac{-1}{2b_1} = 0\]
Hashtags
#orthogonal
#curve
#conicsection
#orthogonal
#analyticgeometry
#mathsolutions
#mathematics
#education
Keywords:::
OrthogonalCurves,
CurveIntersection,
ConicSections,
OrthogonalCondition,
AnalyticGeometry,
Equations,
Question 13,
IntersectionPoints,
OrthogonalTangents,
GeometricConditions,
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