Question

If the curves $a{x}^2+b{y}^2=1$ and $a_1{x}^2+b_1{y}^2=1$ intersect orthogonally then:

• A. $\frac{1}{a}-\frac{1}{b}=\frac{1}{a_1}-\frac{1}{b_1}$
• B. $\frac{1}{a}+\frac{1}{b}=\frac{1}{a_1}+\frac{1}{b_1}$
• C. $\frac{1}{a}-\frac{1}{b}=\frac{1}{a_1}+\frac{1}{b_1}$
• D. None of the above

Answer 1

The correct option is A $\frac{1}{a}-\frac{1}{b}=\frac{1}{a_1}-\frac{1}{b_1}$

If curves $a{x}^2+b{y}^2=1$ and $a_1{x}^2+b_1{y}^2=1$ intersects orthogonally then :

$a{x}^2+b{y}^2=\mathrm{1..........}\left(1\right)a_1{x}^2+b_1{y}^2=\mathrm{1.........}\left(2\right)$

Solve (1) and (2) by multiplying it (1) by $a_1$ and (2) by a and subtracting

$x^2=\frac{b_1-b}{{ab}_1-a_{1}b}{y}^2=\frac{a_1-a}{a_{1}b-{ab}_1}$

Orthogonally intersect $m_1{m}_2=-1$

$2by\frac{dy}{dt}=-2ax{m}_1=-\frac{ax}{by}{m}_2=-\frac{a_{1}x}{b_{1}y}{m}_1{m}_2=\frac{a{a}_1{x}^2}{b{b}_1{y}^2}=-1=\frac{a{a}_1\left(\frac{b-b_1}{{ab}_1-a_{1}b}\right)}{b{b}_1\left(\frac{a_1-a}{a_{1}b-a{b}_1}\right)}=-1=\frac{a-a_1}{a{a}_1}=\frac{b-b_1}{b{b}_1}\frac{1}{a_1}-\frac{1}{a}=\frac{1}{b_1}-\frac{1}{b}\frac{1}{a}-\frac{1}{b}=\frac{1}{a_1}-\frac{1}{b_1}$