Question 114 of 212MediumMCQ
If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :
Solutions · 1
Official solution
When two curves intersect each other at right angle, then at the point of intersection the product of tangent of slopes = −1.
Let m1, and m2 are the tangent of the slope of the two curves respectively
\[\therefore\,\,\,\] m1 m2 = - 1.
Now let they intersect at point (x1, y1)
\[\therefore\,\,\,\] y12=6x, and \[9x_1^2 + b\,y_1^2 = 16\]
y2 = 6x
\[ \Rightarrow \,\,\,\,2y{{dy} \over {dx}} = 6\]
\[ \Rightarrow \,\,\,\,{{dy} \over {dx}} = {3 \over y}\]
\[\therefore\,\,\,\] \[{\left( {{{dy} \over {dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = {3 \over {{y_1}}} = {m_1}\]
9x2 + by2 = 16
\[ = 18x + 2by{{dy} \over {dx}} = O\]
\[ \Rightarrow \,\,\,\,{\left( {{{dy} \over {dx}}} \right)_{({x_1},{y_1})}} = - {{9{x_1}} \over {b{y_1}}} = {m_2}\]
As m1 m2 = -1
\[\therefore\,\,\,\] \[{3 \over {y{}_1}} \times - {{9{x_1}} \over {b{y_1}}} = - 1\]
\[ \Rightarrow \,\,\,\,27{x_1} = by_1^2\]
\[ \Rightarrow \,\,\,\,\,27{x_1} = b.6{x_1}\] \[\,\,\,\] [as \[y_1^2 = 6{x_1}\,]\]
\[ \Rightarrow \,\,\,\,b = {{27} \over 6} = {9 \over 2}\] Answer the question to unlock