Question

- If the curves \[x = {y^4}\] and \[xy = k\] cut at right angles, then find the value of \[{\left( {4k} \right)^6}\].

Answer 1

- Hint: The two given curves cut at right angles. So, the curves have a point of intersection at which a perpendicular common tangent exists. Find the point of intersection and the slopes of the tangent lines. The two tangents being perpendicular, apply the condition of perpendicularity of two straight lines and find the value of \[k\].

Formula used:
If two straight lines having slopes \[{m_1}\] and \[{m_2}\] be perpendicular, then the product of the slopes is equal to \[{m_1}{m_2} = - 1\]
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
\[\dfrac{d}{{dx}}\left\{ {f\left( x \right)g\left( x \right)} \right\} = f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\} + g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}\]
\[\dfrac{d}{{dx}}\left( k \right) = 0\], \[k\] being a constant
\[{a^m} \cdot {a^n} = {a^{m + n}}\]
\[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
\[{\left( {{a^m}} \right)^n} = {a^{mn}}\]

Complete step by step solution:
The given two equations are \[x = {y^4}.....\left( i \right)\] and \[xy = k.....\left( {ii} \right)\]
Solve the equations and find the point of intersection.
Substitute the expression for \[x\] from equation \[\left( i \right)\] in equation \[\left( {ii} \right)\]
\[{y^4} \cdot y = k\]
Find the value of \[y\].
Use the formula \[{a^m} \cdot {a^n} = {a^{m + n}}\]
\[\begin{array}{l} \Rightarrow {y^5} = k\\ \Rightarrow y = {k^{\dfrac{1}{5}}}\end{array}\]
Now, find the value of \[x\] using the value of \[y\].
Putting the value of \[y\] in equation \[\left( i \right)\], we get
\[x = {\left( {{k^{\dfrac{1}{5}}}} \right)^4}\]
Use the formula \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[ \Rightarrow x = {k^{\dfrac{4}{5}}}\]
\[\therefore \] The point of intersection is \[\left( {{k^{\dfrac{4}{5}}},{k^{\dfrac{1}{5}}}} \right)\]
Find the slope of the curves \[\left( i \right)\] and \[\left( {ii} \right)\].
Differentiate both sides of the equation \[\left( i \right)\] with respect to \[y\]
\[\dfrac{d}{{dy}}\left( x \right) = \dfrac{d}{{dy}}\left( {{y^4}} \right).....\left( {iii} \right)\]
Use the formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Putting \[n = 4\], we get \[\dfrac{d}{{dx}}\left( {{x^4}} \right) = 4{x^3}\]
So, \[\dfrac{d}{{dy}}\left( {{y^4}} \right) = 4{y^3}\] and \[\dfrac{d}{{dy}}\left( x \right) = \dfrac{{dx}}{{dy}}\]
\[\therefore \] from equation \[\left( {iii} \right)\], we get \[\dfrac{{dx}}{{dy}} = 4{y^3}\]
Take the reciprocal
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{4{y^3}}}\]
Replace \[x\] by \[{k^{\dfrac{4}{5}}}\] and \[y\] by \[{k^{\dfrac{1}{5}}}\]
\[ \Rightarrow {\left[ {\dfrac{{dy}}{{dx}}} \right]_{\left( {{k^{\dfrac{4}{5}}},{k^{\dfrac{1}{5}}}} \right)}} = \dfrac{1}{{4 \times {{\left( {{k^{\dfrac{1}{5}}}} \right)}^3}}} = \dfrac{1}{{4{k^{\dfrac{3}{5}}}}}\]
So, the slope of the tangent to the curve \[\left( i \right)\] at the point \[\left( {{k^{\dfrac{4}{5}}},{k^{\dfrac{1}{5}}}} \right)\] is \[\dfrac{1}{{4{k^{\dfrac{3}{5}}}}}\]
Differentiate both sides of the equation \[\left( {ii} \right)\] with respect to \[x\]
\[\dfrac{d}{{dx}}\left( {xy} \right) = \dfrac{d}{{dx}}\left( k \right).....\left( {iv} \right)\]
Use the product rule \[\dfrac{d}{{dx}}\left\{ {f\left( x \right)g\left( x \right)} \right\} = f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\} + g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}\]
Putting \[f\left( x \right) = x\] and \[g\left( x \right) = y\], we get
\[\dfrac{d}{{dx}}\left( {xy} \right) = x\dfrac{d}{{dx}}\left( y \right) + y\dfrac{d}{{dx}}\left( x \right)\]
Now, \[\dfrac{d}{{dx}}\left( y \right) = \dfrac{{dy}}{{dx}}\] and \[\dfrac{d}{{dx}}\left( x \right) = 1\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {xy} \right) = x\dfrac{{dy}}{{dx}} + y\]
and \[\dfrac{d}{{dx}}\left( k \right) = 0\], \[k\] being a constant.
From equation \[\left( {iv} \right)\], we get
\[x\dfrac{{dy}}{{dx}} + y = 0\]
Solve this equation for \[\dfrac{{dy}}{{dx}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{y}{x}\]
Replace \[x\] by \[{k^{\dfrac{4}{5}}}\] and \[y\] by \[{k^{\dfrac{1}{5}}}\]
\[ \Rightarrow {\left[ {\dfrac{{dy}}{{dx}}} \right]_{\left( {{k^{\dfrac{4}{5}}},{k^{\dfrac{1}{5}}}} \right)}} = - \dfrac{{{k^{\dfrac{1}{5}}}}}{{{k^{\dfrac{4}{5}}}}}\]
Use the formula \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
\[ \Rightarrow {\left[ {\dfrac{{dy}}{{dx}}} \right]_{\left( {{k^{\dfrac{4}{5}}},{k^{\dfrac{1}{5}}}} \right)}} = - {k^{\dfrac{1}{5} - \dfrac{4}{5}}} = - {k^{ - \dfrac{3}{5}}}\]
So, the slope of the tangent to the curve \[\left( {ii} \right)\] at the point \[\left( {{k^{\dfrac{4}{5}}},{k^{\dfrac{1}{5}}}} \right)\] is \[ - {k^{ - \dfrac{3}{5}}}\]
Use the condition for perpendicularity of two straight lines.
Product of slopes of two perpendicular straight lines is equal to \[\left( { - 1} \right)\].
So,
\[\begin{array}{l}\left( {\dfrac{1}{{4{k^{\dfrac{3}{5}}}}}} \right)\left( { - {k^{ - \dfrac{3}{5}}}} \right) = - 1\\ \Rightarrow - \dfrac{{{k^{ - \dfrac{3}{5}}}}}{{4{k^{\dfrac{3}{5}}}}} = - 1\\ \Rightarrow \dfrac{1}{4}{k^{ - \dfrac{3}{5} - \dfrac{3}{5}}} = 1\\ \Rightarrow {k^{ - \dfrac{6}{5}}} = 4\end{array}\]
Multiplying the power of both sides by \[\left( { - \dfrac{5}{6}} \right)\]
\[ \Rightarrow {\left( {{k^{ - \dfrac{6}{5}}}} \right)^{ - \dfrac{5}{6}}} = {\left( 4 \right)^{ - \dfrac{5}{6}}}\]
Use the formula \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[ \Rightarrow k = {\left( 4 \right)^{ - \dfrac{5}{6}}}\]
Multiply both sides by \[4\]
\[ \Rightarrow 4k = 4 \times {\left( 4 \right)^{ - \dfrac{5}{6}}}\]
Use the formula \[{a^m} \cdot {a^n} = {a^{m + n}}\]
\[ \Rightarrow 4k = {4^{1 - \dfrac{5}{6}}} = {4^{\dfrac{1}{6}}}\]
Multiply the power of both sides by \[6\]
\[ \Rightarrow {\left( {4k} \right)^6} = {\left( {{4^{\dfrac{1}{6}}}} \right)^6}\]
Use the formula \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[ \Rightarrow {\left( {4k} \right)^6} = 4\]

Hence the value of \[{\left( {4k} \right)^6}\] is \[4\].

Note: If two straight lines are parallel, then the slopes of the lines are proportional but if the lines be perpendicular, then product of slopes is equal to \[\left( { - 1} \right)\]. Differentiating a function \[y = f\left( x \right)\] with respect to \[x\] at a point, we get slope of that curve at that point.