Question

If the curve ${y^2} = 6x$ and $9{x^2} + b{y^2} = 16$ intersect each other at right angles, then find the value of $b$
(A) $4$
(B) $\dfrac{9}{2}$
(C) $6$
(D) $\dfrac{7}{2}$

Answer 1

Hint: Try to understand the question well, before starting the solution. First, find the slope for both the curves by differentiating the equations. The product of slopes at a right angle is $ - 1$, use this information to establish an expression for $b$. Now, use the given equation in the expression of $b$ to find the exact value.

Complete step-by-step answer:
In the question, we are given an equation of two curves and they intersect each other at a right angle. So, if these curves intersect each other then there must exist one value of x and y that satisfies both of them.
So, we have:
${y^2} = 6x$........................(1)
$9{x^2} + b{y^2} = 16$........................(2)
As we know that the first derivative of a curve gives us the equation of the slope of the curve. So, we will derive the equation (1) and (2) to find their slopes:
$ \Rightarrow \dfrac{{d\left( {{y^2}} \right)}}{{dx}} = \dfrac{{d\left( {6x} \right)}}{{dx}} \Rightarrow 2y\dfrac{{dy}}{{dx}} = 6$
Let’s name the slope of equation (1) as ${m_1}$, so we get:
$ \Rightarrow \dfrac{{dy}}{{dx}} = {m_1} = \dfrac{6}{{2y}} = \dfrac{3}{y}$ ………………….(3)
Similarly for equation (2), by taking derivative on both sides:
$ \Rightarrow \dfrac{{d\left( {9{x^2} + b{y^2}} \right)}}{{dx}} = \dfrac{{d\left( {16} \right)}}{{dx}} \Rightarrow 18x + 2by\dfrac{{dy}}{{dx}} = 0$
Now let’s name the slope of equation (2) as ${m_2}$, so we get:
$ \Rightarrow \dfrac{{dy}}{{dx}} = {m_2} = \dfrac{{ - 18x}}{{2by}} = \dfrac{{ - 9x}}{{by}}$ …………………..(4)
Since we know that the product of the slopes of the two perpendicularly intersecting curves is $ - 1$. And we are given that curve (1) and (2) are intersecting at a right angle
$ \Rightarrow {m_1} \times {m_2} = - 1$
Now we can substitute the values of ${m_1}$ and ${m_2}$ from equation (3) and (4)
$ \Rightarrow {m_1} \times {m_2} = \dfrac{3}{y} \times \dfrac{{ - 9x}}{{by}} = - 1$
We can find the expression for the value of b from the above equation
$ \Rightarrow \dfrac{3}{y} \times \dfrac{{ - 9x}}{{by}} = - 1 \Rightarrow b = \dfrac{{27x}}{{{y^2}}}$
So, we got the value of $b$ as $\dfrac{{27x}}{{{y^2}}}$. At this point, we can use the equation (1) to express the denominator term in terms of x
$ \Rightarrow b = \dfrac{{27x}}{{{y^2}}} = \dfrac{{27x}}{{6x}} = \dfrac{9}{2}$
Hence, we get the value of $b = \dfrac{9}{2}$

Thus, the option (B) is the correct answer

Note: Notice that the concept of differentiation played a crucial role while solving this problem. When we derive any curve, we get the equation of the slope of that curve, i.e. $\dfrac{{dy}}{{dx}}$. And the property that the product of two slopes who are perpendicular to each other is $ - 1$. Two lines are parallel if their slopes are equal.