Question

Show that the two curves $a{x^2} + b{y^2} = 1$ and $c{x^2} + d{y^2} = 1$ cut each other orthogonally, if $\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{c} - \dfrac{1}{d}$.

Answer 1

Here, we need to prove that the two curves cut orthogonally. We will assume that the given curves cut each other at a particular point. Since this point lies on both the curves, it satisfies the two equations. We will substitute the point into the equations of the curves, and subtract the resulting equations from each other. Then, we will simplify the given condition to get another equation. Now, we will find the slopes of the two curves by differentiating the equations and substituting the coordinates of the point of intersection. Finally, we will multiply the slopes of the two curves to get the result as $- 1$. Two curves cut each other orthogonally if the product of their slopes is equal to $- 1$.

Complete step by step solution:
Let the curves $a{x^2} + b{y^2} = 1$ and $c{x^2} + d{y^2} = 1$ cut each other at the point $\left( {{x_1},{y_1}} \right)$.
We know that the point $\left( {{x_1},{y_1}} \right)$ will lie on both the curves.
Therefore, the point will satisfy the equations of both the curves.
Substituting $x = {x_1}$ and $y = {y_1}$ in the curves $a{x^2} + b{y^2} = 1$ and $c{x^2} + d{y^2} = 1$, we get
$ax_1^2 + by_1^2 = 1 \ldots \ldots \ldots \left( 1 \right)$
$cx_1^2 + dy_1^2 = 1 \ldots \ldots \ldots \left( 2 \right)$
Now, we will subtract equation $\left( 2 \right)$ from equation $\left( 1 \right)$.
Subtracting equation $\left( 2 \right)$ from equation $\left( 1 \right)$, we get
$\left( {ax_1^2 + by_1^2} \right) - \left( {cx_1^2 + dy_1^2} \right) = 1 - 1$
Simplifying the equation, we get
$\Rightarrow ax_1^2 + by_1^2 - cx_1^2 - dy_1^2 = 0$
Factoring out $x_1^2$ and $y_1^2$, we get
$\Rightarrow \left( {a - c} \right)x_1^2 + \left( {b - d} \right)y_1^2 = 0$
Rewriting the equation, we get
$\begin{array}{l} \Rightarrow \left( {a - c} \right)x_1^2 = - \left( {b - d} \right)y_1^2\\\ \Rightarrow \dfrac{{a - c}}{{b - d}} = - \dfrac{{y_1^2}}{{x_1^2}} \ldots \ldots \ldots \left( 3 \right)\end{array}$
Now, it is given that $\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{c} - \dfrac{1}{d}$.
Rewriting the equation, we get
$\begin{array}{l}\dfrac{1}{a} - \dfrac{1}{c} = \dfrac{1}{b} - \dfrac{1}{d}\\\ \Rightarrow \dfrac{1}{c} - \dfrac{1}{a} = \dfrac{1}{d} - \dfrac{1}{b}\end{array}$
Taking L.C.M. on both sides of the equation, we get
$\Rightarrow \dfrac{{a - c}}{{ac}} = \dfrac{{b - d}}{{bd}}$
Rewriting the equation, we get
$\Rightarrow \dfrac{{a - c}}{{b - d}} = \dfrac{{ac}}{{bd}} \ldots \ldots \ldots \left( 4 \right)$
From equation $\left( 3 \right)$ and equation $\left( 4 \right)$, we get
$\dfrac{{ac}}{{bd}} = - \dfrac{{y_1^2}}{{x_1^2}}$
Now, we will find the slopes of the two curves at the point of intersection $\left( {{x_1},{y_1}} \right)$.
The slope of the curves $a{x^2} + b{y^2} = 1$ and $c{x^2} + d{y^2} = 1$ at the point $\left( {{x_1},{y_1}} \right)$ can be calculated by differentiating the curves and substituting $x = {x_1}$ and $y = {y_1}$.
Differentiating both sides of the curve $a{x^2} + b{y^2} = 1$, we get
$2ax + 2byy' = 0$
Simplifying the equation, we get
$\begin{array}{l} \Rightarrow 2byy' = - 2ax\\\ \Rightarrow byy' = - ax\\\ \Rightarrow y' = - \dfrac{{ax}}{{by}}\end{array}$
Substituting $x = {x_1}$ and $y = {y_1}$, we get
$\Rightarrow y' = - \dfrac{{a{x_1}}}{{b{y_1}}}$
Similarly, differentiating both sides of the curve $c{x^2} + d{y^2} = 1$, we get
$2cx + 2dyy' = 0$
Simplifying the equation, we get
$\begin{array}{l} \Rightarrow 2dyy' = - 2cx\\\ \Rightarrow dyy' = - cx\\\ \Rightarrow y' = - \dfrac{{cx}}{{dy}}\end{array}$
Substituting $x = {x_1}$ and $y = {y_1}$, we get
$\Rightarrow y' = - \dfrac{{c{x_1}}}{{d{y_1}}}$
Now, we will prove that the two curves cut each other orthogonally at $\left( {{x_1},{y_1}} \right)$.
Two curves cut each other orthogonally if the product of their slopes is equal to $- 1$.
Multiplying the slopes of the curves $a{x^2} + b{y^2} = 1$ and $c{x^2} + d{y^2} = 1$, we get
$\left( { - \dfrac{{a{x_1}}}{{b{y_1}}}} \right)\left( { - \dfrac{{c{x_1}}}{{d{y_1}}}} \right) = \dfrac{{acx_1^2}}{{bdy_1^2}}$
Rewriting the equation, we get
$\Rightarrow \left( { - \dfrac{{a{x_1}}}{{b{y_1}}}} \right)\left( { - \dfrac{{c{x_1}}}{{d{y_1}}}} \right) = \dfrac{{ac}}{{bd}} \times \dfrac{{x_1^2}}{{y_1^2}}$
Substituting $\dfrac{{ac}}{{bd}} = - \dfrac{{y_1^2}}{{x_1^2}}$ in the equation, we get
$\Rightarrow \left( { - \dfrac{{a{x_1}}}{{b{y_1}}}} \right)\left( { - \dfrac{{c{x_1}}}{{d{y_1}}}} \right) = - \dfrac{{y_1^2}}{{x_1^2}} \times \dfrac{{x_1^2}}{{y_1^2}}$
Simplifying the expression, we get
$\Rightarrow \left( { - \dfrac{{a{x_1}}}{{b{y_1}}}} \right)\left( { - \dfrac{{c{x_1}}}{{d{y_1}}}} \right) = - 1$

Therefore, since the product of the slopes of the curves $a{x^2} + b{y^2} = 1$ and $c{x^2} + d{y^2} = 1$ at the point $\left( {{x_1},{y_1}} \right)$ is equal to $- 1$, the curves cut each other orthogonally.

Note:
We have used $y'$ to denote the slope of the lines instead of $\dfrac{{dy}}{{dx}}$, to avoid the confusion between $dy$ in $\dfrac{{dy}}{{dx}}$, and the product of $d$ and $y$. Also, remember that two curves cut each other orthogonally at a point if the product of their slopes at the point of intersection is equal to $- 1$. A common mistake is to equate the product of their slopes to 1, which will give us the wrong answer.