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If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :

JEE · Math · previous-year question

  1. A.$${9 \over 2}$$correct
  2. B.6
  3. C.$${7 \over 2}$$
  4. D.4

Answer

A. $${9 \over 2}$$

Explanation

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) $$\therefore\,\,\,$$ $$y_1^2 = 6x,$$ 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 $$y_1^2 = 6{x_1}\,]$$ $$ \Rightarrow \,\,\,\,b = {{27} \over 6} = {9 \over 2}$$

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