Question

Assertion :Two curves $a{x}^2+b{y}^2=1$ & $a^{\prime }{x}^2+b^{\prime }{y}^2=1$ are orthogonal if $\frac{1}{a}-\frac{1}{b}=\frac{1}{a^{\prime }}-\frac{1}{b^{\prime }}$ Reason: Two curves intersect orthogonally at a point if product of their slopes at that point is $-1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Given curves are $a{x}^2+b{y}^2=1$ (i)
& $a^{\prime }{x}^2+b^{\prime }{y}^2=1$ (ii)
Let $(x_1,y_1)$ be the point of intersection of the curves then
$a{x}_1^2+b{y}_1^2=1$ (iii)
& $a^{\prime }{x}_1^2+b^{\prime }{y}_1^2=1$ (iv)
Differentiating (i) and (ii) w.r.t to $x$
$\therefore$ $2ax+2by\frac{dy}{dx}=0$ & $2a^{\prime }x+2b^{\prime }y\frac{dy}{dx}=0$
$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=m_1=-\frac{a{x}_1}{b{y}_1}$ & ${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=m_2=-\frac{a^{\prime }{x}_1}{b^{\prime }{y}_1}$
Now $m_1{m}_2=-1$
$\Rightarrow$ $(-\frac{a{x}_1}{b{y}_1})(-\frac{a^{\prime }{x}_1}{b^{\prime }{y}_1})=-1$
$\Rightarrow$ $a{a}^{\prime }{x_1}^2=-b{b}^{\prime }{y_1}^2$ (v)
and from given curves on subtracting iii) and iv), we get
$(a-a^{\prime }){x_1}^2=-(b-b^{\prime }){y_1}^2$ (vi)
On dividing (vi) by (v) we get
$\frac{a-a^{\prime }}{a{a}^{\prime }}=\frac{b-b^{\prime }}{b{b}^{\prime }}$ or $\frac{a^{\prime }-a}{a{a}^{\prime }}=\frac{b^{\prime }-b}{b{b}^{\prime }}$

$\Rightarrow$ $\frac{1}{a}-\frac{1}{b}=\frac{1}{a^{\prime }}-\frac{1}{b^{\prime }}$