Question

The value of m for which the area of the triangle included between the axes and any other tangent to the ${{x}^{m}}y={{b}^{m}}$ is constant, is
(A) $\dfrac{1}{2}$
(B) 1
(C) $\dfrac{3}{2}$
(D) 2

Answer 1

Assume that the tangent to the curve ${{x}^{m}}y={{b}^{m}}$ is touching a point P $\left( {{x}_{1}},{{y}_{1}} \right)$ on the curve. Now, differentiate the equation of the curve with respect to $dx$ and calculate the slope that is $\dfrac{dy}{dx}$ . Using the coordinate of point P, get the slope at point P. We also know the standard equation of a straight line having its slope equal to $m$ and passing through a point $\left( {{x}_{1}},{{y}_{1}} \right)$ , $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ . Use this property and get the equation of the tangent passing through point P $\left( {{x}_{1}},{{y}_{1}} \right)$ . Now, calculate the y-intercept of the tangent by putting $x=0$ in the equation of the tangent. Similarly, calculate the x-intercept of the tangent by putting $y=0$ in the equation of the tangent. Solve it further and calculate the area of $\Delta OAB$ by using the formula, Area = $\dfrac{1}{2}\times Base\times Perpendicular$ . Now, use the relation ${{y}_{1}}=\dfrac{{{b}^{m}}}{{{x}_{1}}^{m}}$ and obtain the area of $\Delta OAB$ in terms of ${{x}_{1}}$ . For the area to be constant the exponent of ${{x}_{1}}$ must be equal to zero. Finally, calculate the value of $m$ .

Complete step by step answer:
According to the question, we are given that,
The equation of the curve is ${{x}^{m}}y={{b}^{m}}$ …………………………………..(1)
We know the property that the slope of a curve is equal to $\dfrac{dy}{dx}$ …………………………………….(2)
Now, on differentiating equation (1) with respect to $dx$ , we get
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{m}}y \right)=\dfrac{d}{dx}\left( {{b}^{m}} \right)$ ……………………………………………….(3)
We know the formula, $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ ……………………………………..(4)
From equation (3) and equation (4), we get
$\Rightarrow {{x}^{m}}\dfrac{d\left( y \right)}{dx}+y\dfrac{d\left( {{x}^{m}} \right)}{dx}=\dfrac{d\left( {{b}^{m}} \right)}{dx}$
$\Rightarrow {{x}^{m}}\dfrac{dy}{dx}+y\times m{{x}^{m-1}}=\dfrac{d\left( {{b}^{m}} \right)}{dx}$ ………………………………………….(5)
We know the property that the differentiation of constant term is equal to zero ………………………………(6)
From equation (5) and equation (6), we get

& \Rightarrow {{x}^{m}}\dfrac{dy}{dx}+y\times m{{x}^{m-1}}=0 \\\ & \Rightarrow {{x}^{m}}\dfrac{dy}{dx}=-y\times m{{x}^{m-1}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-y\times m{{x}^{m-1}}}{{{x}^{m}}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-ym}{{{x}^{m-\left( m-1 \right)}}} \\\ \end{aligned}$$ $$\Rightarrow \dfrac{dy}{dx}=\dfrac{-my}{x}$$ ………………………………..(7) On solving equation (1), we get $$\Rightarrow {{x}^{m}}y={{b}^{m}}$$ $$\Rightarrow y=\dfrac{{{b}^{m}}}{{{x}^{m}}}$$ …………………………………………………(8) Now, from equation (7) and equation (8), we get $$\Rightarrow \dfrac{dy}{dx}=\dfrac{-m}{x}\times \dfrac{{{b}^{m}}}{{{x}^{m}}}$$ $$\Rightarrow \dfrac{dy}{dx}=\dfrac{-m{{b}^{m}}}{{{x}^{m+1}}}$$ …………………………………………(9) Here, let us assume the tangent to the curve $${{x}^{m}}y={{b}^{m}}$$ is touching a point P $$\left( {{x}_{1}},{{y}_{1}} \right)$$ on the curve. From equation (9), we have an equation for the slope of a tangent. At point P $$\left( {{x}_{1}},{{y}_{1}} \right)$$ , we have The slope of the tangent = $$\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}$$ ………………………………….(10) We also know the standard equation of a straight line having its slope equal to $$m$$ and passing through a point $$\left( {{x}_{1}},{{y}_{1}} \right)$$ , $$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$$ ………………………………(11) The tangent is passing through point P $$\left( {{x}_{1}},{{y}_{1}} \right)$$ ……………………………………..(12) Now, from equation (10), equation (11), and equation (12), we get The equation of the tangent to the curve, $$\left( y-{{y}_{1}} \right)=\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}\left( x-{{x}_{1}} \right)$$ ………………………………………………..(13) We also know the property that when a straight line is passing through the y-axis then at the point of intersection, the x coordinate is equal to zero ………………………………………(14) Using the property shown in equation (14) and on putting $$x=0$$ in equation (13), we get $$\begin{aligned} & \Rightarrow \left( y-{{y}_{1}} \right)=\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}\left( 0-{{x}_{1}} \right) \\\ & \Rightarrow \left( y-{{y}_{1}} \right)=\dfrac{m{{b}^{m}}}{{{x}_{1}}^{m+1}}\times {{x}_{1}} \\\ & \Rightarrow y=\dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+{{y}_{1}} \\\ \end{aligned}$$ From the above equation, we have calculated the y coordinate of the point of intersection of the tangent and y-axis. So, the coordinate of the point of intersection of tangent and y-axis is A $$\left( 0,\dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+{{y}_{1}} \right)$$ ………………………………………….(15) Similarly, we also know the property that when a straight line is passing through the x-axis then at the point of intersection, the y coordinate is equal to zero ………………………………………(16) Using the property shown in equation (16) and on putting $$y=0$$ in equation (13), we get $$\begin{aligned} & \Rightarrow \left( 0-{{y}_{1}} \right)=\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}\left( x-{{x}_{1}} \right) \\\ & \Rightarrow \dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}=\left( x-{{x}_{1}} \right) \\\ & \Rightarrow x=\dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}+{{x}_{1}} \\\ \end{aligned}$$ From the above equation, we have calculated the x coordinate of the point of intersection of the tangent and y-axis. So, the coordinate of the point of intersection of tangent and y-axis is B $$\left( \dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}+{{x}_{1}},0 \right)$$ ………………………………………….(17) Now, on plotting these all on the coordinate axes, we get  From the above diagram, we can observe that, $$\Delta OAB$$ is the required triangle and $$\Delta OAB$$ is right-angled at O where O is the origin. The length of the base of $$\Delta OAB$$ , OB = $$\left( \dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}+{{x}_{1}} \right)$$ ………………………………………………..(18) The length of perpendicular of $$\Delta OAB$$ , OA = $$\left( \dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+{{y}_{1}} \right)$$ ………………………………………………..(19) We also know the formula for the area of the triangle, Area = $$\dfrac{1}{2}\times Base\times Perpendicular$$ ………………………………………(20) Now, from equation (18), equation (19), and equation (20), we get The area of $$\Delta OAB$$ $$=\dfrac{1}{2}\times \left( \dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}+{{x}_{1}} \right)\times \left( \dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+{{y}_{1}} \right)$$ $$=\dfrac{1}{2}\times \left( \dfrac{{{y}_{1}}{{x}_{1}}^{m}\times {{x}_{1}}}{m{{b}^{m}}}+{{x}_{1}} \right)\left( \dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+{{y}_{1}} \right)$$ ……………………………………..(21) Since the point P $$\left( {{x}_{1}},{{y}_{1}} \right)$$ is on the curve so, the coordinates of point P must satisfy the equation of the curve. Putting $$x={{x}_{1}}$$ and $$y={{y}_{1}}$$ in equation (1), we get $$\Rightarrow {{x}_{1}}^{m}{{y}_{1}}={{b}^{m}}$$ ……………………………………………….(22) $$\Rightarrow {{y}_{1}}=\dfrac{{{b}^{m}}}{{{x}_{1}}^{m}}$$ ………………………………………..(23) Now, from equation (21), equation (22), and equation (23), we get The area of $$\Delta OAB$$ $$\begin{aligned} & =\dfrac{1}{2}\times \left( \dfrac{{{b}^{m}}\times {{x}_{1}}}{m{{b}^{m}}}+{{x}_{1}} \right)\left( \dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+\dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right) \\\ & =\dfrac{1}{2}\times \left( \dfrac{{{x}_{1}}+m{{x}_{1}}}{m} \right)\left( \dfrac{m{{b}^{m}}+{{b}^{m}}}{{{x}_{1}}^{m}} \right) \\\ & =\dfrac{1}{2}\times {{x}_{1}}\left( \dfrac{m+1}{m} \right)\left( \dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right)\left( m+1 \right) \\\ & =\dfrac{1}{2}\times \dfrac{{{\left( m+1 \right)}^{2}}}{m}\left( \dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right){{x}_{1}} \\\ & =\dfrac{1}{2}\times \dfrac{{{\left( m+1 \right)}^{2}}}{m}\times {{b}^{m}}\times {{x}_{1}}^{\left( 1-m \right)} \\\ \end{aligned}$$ Since the area of the triangle is a constant so, in the above equation the term $${{x}_{1}}^{\left( 1-m \right)}$$ must also be constant and it will be constant if and only if the term $$\left( 1-m \right)$$ is equal to zero. That is, $$\begin{aligned} & \Rightarrow 1-m=0 \\\ & \Rightarrow m=1 \\\ \end{aligned}$$ Therefore, the value of m is equal to 1. **So, the correct answer is “Option B”.** **Note:** To solve these types of questions we have to take a few points into our consideration. That is, the slope of a curve is always equal to $$\dfrac{dy}{dx}$$ . The other point is that when a straight line is passing through the y-axis then at the point of intersection, the x coordinate is equal to zero. Similarly, when a straight line is passing through the x-axis then at the point of intersection, the y coordinate is equal to zero.