If the curves x^2a+y^2b=1 and x^2c+y^2d=1 intersect each other at an angle of 90^, then wh…

Question

If the curves $\dfrac{x^2}a+\dfrac{y^2}b=1$ and $\dfrac{x^2}c+\dfrac{y^2}d=1$ intersect each other at an angle of $90^\circ$, then which of the following relations is true?

Accepted Answer

$${{{x^2}} \over a} + {{{y^2}} \over b} = 1$$ ..........(1)

Differentiating both sides :

$${{2x} \over a} + {{2y} \over b}{{dy} \over {dx}} = 0 \Rightarrow {y \over b}{{dy} \over {dx}} = {{ - x} \over a}$$

$$ \Rightarrow $$ $${{dy} \over {dx}} = {{ - bx} \over {ay}}$$ ............(2)

$${{{x^2}} \over c} + {{{y^2}} \over d} = 1$$ ........(3)

Differentiating both sides : $${{dy} \over {dx}} = {{ - dx} \over {cy}}$$ ...........(4)

$${m_1}{m_2} =  - 1 \Rightarrow {{ - bx} \over {ay}} \times {{ - dx} \over {cy}} =  - 1$$

$$ \Rightarrow bd{x^2} =  - ac{y^2}$$ .............(5)

$$(1) - (3) \Rightarrow \left( {{1 \over a} - {1 \over c}} \right){x^2} + \left( {{1 \over b} - {1 \over d}} \right){y^2} = 0$$

$$ \Rightarrow {{c - a} \over {ac}}{x^2} + {{d - b} \over {bd}} \times \left( {{{ - bd} \over {ac}}} \right){x^2} = 0$$ (using 5)

$$ \Rightarrow (c - a) - (d - b) = 0$$

$$ \Rightarrow c - a = d - b$$

$$ \Rightarrow c - d = a - b$$

Solutions · 1