show that the condition that the curves ax2+by2=1 and a'x2+b'y2=1 should intersect orthogonally (at900) such that 1/a-1/b=1/a'-1/b' - Maths - Application of Derivatives

Question

show that the condition that the curves
ax2+by2=1 and
a'x2+b'y2=1 should intersect orthogonally
(at900) such that 1/a-1/b=1/a'-1/b'

brainmasswriter1 · Accepted Answer

Let the point of intersection of the curves be $(x_{1}, y_{1})$ , so this point must satisfy both the equations

So now equations become as shown below:

$a {x_{1}}^{2} + {b y_{1}}^{2} = 1 ... (1) a' {x_{1}}^{2} + {b' y_{1}}^{2} = 1 ... (2)$

Solving (1) and (2) for ${x_{1}}^{2} a n d {y_{1}}^{2}$ we get:

${x_{1}}^{2} = \frac{b' - b}{a b' - a b'} a n d {y_{1}}^{2} = - \frac{a' - a}{a b' - a b'} ... (3)$

Differentiating the given equations w.r.t x we get:

$\frac{d y}{d x} = \frac{- a x}{a y} a n d \frac{d y}{d x} = \frac{- a' x}{b' y}$

At point $(x_{1}, y_{1})$ the slopes would be:

$\frac{d y}{d x} = \frac{- a x_{1}}{b y_{1}} a n d \frac{d y}{d x} = \frac{- a' x_{1}}{b' y_{1}}$

Now the curves intersect orthogonally at point $(x_{1}, y_{1})$ if the product of slopes=-1, so it means:

$\frac{- a x_{1}}{b y_{1}} \times \frac{- a' x_{1}}{b' y_{1}} = - 1 \Rightarrow \frac{a a'}{b b'} \times \frac{{x_{1}}^{2}}{{y_{1}}^{2}} = - 1 \Rightarrow \frac{a a'}{b b'} (- \frac{b' - b}{a' - a}) = - 1 \Rightarrow \frac{b' - b}{b b'} = \frac{a' - a}{a a'} \Rightarrow \frac{1}{b} - \frac{1}{b'} = \frac{1}{a} - \frac{1}{a'} \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{a'} - \frac{1}{b'}$