Question

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

${\sf{a)}} \: \: \frac{9}{2}$
${\sf{b) \: \: 6}}$
${\sf{b) \: 4}}$
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Answer 1

$\small \bold{It \: is \: given \: that \: the \: curves y {}^{2} =6x and}$

$\small \bold{9x^2+by^2 =16 \: intersect \: at \: right \: angles.}$

$\small \bold{Let \: the \: slope \: of \: y^2=6x \: be \: m_1}$

Also, let the slope of 9x² + by² = 16 be$\bold {m_2}$

$\small \bold{Thus, we \: have \: m_1 \: m_2 =−1}$

Differentiating both the equations wrt x we get

$\bold{2y\frac{dy}{dx} = 6}$

$\bold{Or, ym_1=3 - - - - - (1)}$

$\bold{Also, 18x+2by \frac{dy}{dx} = 0}$

$\bold{Or, m^2 = \frac{ - 9x}{by}}$

$\bold{We \: know \: that \: m_1 m_2 =−1}$

$\bold{Thus, \: \frac{3}{y} \times \frac{ - 9x}{by} = 1}$

$\bold{ \rightarrow \: b = \frac{27x}{ {y}^{2} } }$

Putting y²=6x in the above equation we get

$\bold{b = \frac{27x}{6x} = \frac{9}{2}}$

Answer 2

Answer:

Y^2 =6x

differentiating, 2ydy/dx=6

dy/dx=3/y

9x^2+b^2y^2=16

differentiating,

18x+2b^2ydy/dx=0

dy/dx=-9x/by

two curves intersect at right angles if product of their slopes or derivatives at point of intersection is -1.

that is (dy/dx)1*(dy/dx)2=-1

(3*-9x)/by^2=-1

y^2=6x

then

-27x/6bx=-1

9/2b=1

2b=9

b= 9/2

Explanation:

Hope it will help you..