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
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Hint: 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
\[\begin{align}
& \Rightarrow {{x}^{m}}\dfrac{dy}{dx}+y\times m{{x}^{m-1}}=0 \\
& \Rightarrow {{x}^{m}}\dfrac{dy}{dx}=-y\times m{{x}^{m-1}} \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{-y\times m{{x}^{m-1}}}{{{x}^{m}}} \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{-ym}{{{x}^{m-\left( m-1 \right)}}} \\
\end{align}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-my}{x}\] ………………………………..(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
\[\begin{align}
& \Rightarrow \left( y-{{y}_{1}} \right)=\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}\left( 0-{{x}_{1}} \right) \\
& \Rightarrow \left( y-{{y}_{1}} \right)=\dfrac{m{{b}^{m}}}{{{x}_{1}}^{m+1}}\times {{x}_{1}} \\
& \Rightarrow y=\dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+{{y}_{1}} \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \left( 0-{{y}_{1}} \right)=\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}\left( x-{{x}_{1}} \right) \\
& \Rightarrow \dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}=\left( x-{{x}_{1}} \right) \\
& \Rightarrow x=\dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}+{{x}_{1}} \\
\end{align}\]
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 \[\Delta OAB\]
\[\begin{align}
& =\dfrac{1}{2}\times \left( \dfrac{{{b}^{m}}\times {{x}_{1}}}{m{{b}^{m}}}+{{x}_{1}} \right)\left( \dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+\dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right) \\
& =\dfrac{1}{2}\times \left( \dfrac{{{x}_{1}}+m{{x}_{1}}}{m} \right)\left( \dfrac{m{{b}^{m}}+{{b}^{m}}}{{{x}_{1}}^{m}} \right) \\
& =\dfrac{1}{2}\times {{x}_{1}}\left( \dfrac{m+1}{m} \right)\left( \dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right)\left( m+1 \right) \\
& =\dfrac{1}{2}\times \dfrac{{{\left( m+1 \right)}^{2}}}{m}\left( \dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right){{x}_{1}} \\
& =\dfrac{1}{2}\times \dfrac{{{\left( m+1 \right)}^{2}}}{m}\times {{b}^{m}}\times {{x}_{1}}^{\left( 1-m \right)} \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow 1-m=0 \\
& \Rightarrow m=1 \\
\end{align}\]
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.
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
\[\begin{align}
& \Rightarrow {{x}^{m}}\dfrac{dy}{dx}+y\times m{{x}^{m-1}}=0 \\
& \Rightarrow {{x}^{m}}\dfrac{dy}{dx}=-y\times m{{x}^{m-1}} \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{-y\times m{{x}^{m-1}}}{{{x}^{m}}} \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{-ym}{{{x}^{m-\left( m-1 \right)}}} \\
\end{align}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-my}{x}\] ………………………………..(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
\[\begin{align}
& \Rightarrow \left( y-{{y}_{1}} \right)=\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}\left( 0-{{x}_{1}} \right) \\
& \Rightarrow \left( y-{{y}_{1}} \right)=\dfrac{m{{b}^{m}}}{{{x}_{1}}^{m+1}}\times {{x}_{1}} \\
& \Rightarrow y=\dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+{{y}_{1}} \\
\end{align}\]
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
\[\begin{align}
& \Rightarrow \left( 0-{{y}_{1}} \right)=\dfrac{-m{{b}^{m}}}{{{x}_{1}}^{m+1}}\left( x-{{x}_{1}} \right) \\
& \Rightarrow \dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}=\left( x-{{x}_{1}} \right) \\
& \Rightarrow x=\dfrac{{{y}_{1}}{{x}_{1}}^{m+1}}{m{{b}^{m}}}+{{x}_{1}} \\
\end{align}\]
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 \[\Delta OAB\]
\[\begin{align}
& =\dfrac{1}{2}\times \left( \dfrac{{{b}^{m}}\times {{x}_{1}}}{m{{b}^{m}}}+{{x}_{1}} \right)\left( \dfrac{m{{b}^{m}}}{{{x}_{1}}^{m}}+\dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right) \\
& =\dfrac{1}{2}\times \left( \dfrac{{{x}_{1}}+m{{x}_{1}}}{m} \right)\left( \dfrac{m{{b}^{m}}+{{b}^{m}}}{{{x}_{1}}^{m}} \right) \\
& =\dfrac{1}{2}\times {{x}_{1}}\left( \dfrac{m+1}{m} \right)\left( \dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right)\left( m+1 \right) \\
& =\dfrac{1}{2}\times \dfrac{{{\left( m+1 \right)}^{2}}}{m}\left( \dfrac{{{b}^{m}}}{{{x}_{1}}^{m}} \right){{x}_{1}} \\
& =\dfrac{1}{2}\times \dfrac{{{\left( m+1 \right)}^{2}}}{m}\times {{b}^{m}}\times {{x}_{1}}^{\left( 1-m \right)} \\
\end{align}\]
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,
\[\begin{align}
& \Rightarrow 1-m=0 \\
& \Rightarrow m=1 \\
\end{align}\]
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.
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