Question

Find the condition that the following conics may cut orthogonally: $a{x}^2+b{y}^2=1$ and $a{x}^2+b^{\prime }{y}^2=1.$

• A. $\frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{b^{\prime }}$
• B. $\frac{1}{a}+\frac{1}{b}=\frac{1}{d}-\frac{1}{b^{\prime }}$
• C. $\frac{1}{a}+\frac{1}{b}=\frac{1}{d}+\frac{1}{b^{\prime }}$
• D. $\frac{1}{a}-\frac{1}{b}=\frac{1}{d}+\frac{1}{b^{\prime }}$

Answer 1

The correct option is A $\frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{b^{\prime }}$

Given, $a{x}^2+b{y}^2=1,a{x}^2+b^{\prime }{y}^2=1$

$\left(\frac{dy}{dx}\right)=-\frac{ax}{by}=m_1,{\left(\frac{dy}{dx}\right)}_{⨿}=-\frac{dx}{b^{\prime }y}=m_2$
If the curves
cut orthogonally, then $m_1{m}_2=-1$ or
$(-\frac{ax}{by})(-\frac{dx}{b^{\prime }y})=-1$ or $ad{x}^2+b{b}^{\prime }{y}^2=0.$ ...(1)
We have
now to put the values of $x^2$ and $y^2$ corresponding to the points of intersection of $a{x}^2+b{y}^2-1=0$ and $a{x}^2+b^{\prime }{y}^2-1=0$
$\therefore \frac{x^2}{b^{\prime }b}=\frac{y^2}{a-d}=\frac{1}{ab-db}$ Putting the values
of $x^2\text{and}{y}^{2}in\left(1\right)$ $ad\frac{(b^{\prime }-b)}{ab-db}+b{b}^{\prime }.\frac{(a-d)}{ab-db}=0$
$\frac{b-b}{bb}+\frac{a-d}{ad}=0$ or $(\frac{1}{b}-\frac{1}{b^{\prime }})+(\frac{1}{d}-\frac{1}{a})=0$ or $\frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{b^{\prime }}$ is the required
condition.