Differential Calculus> see attachment and explain it...

see attachment and explain it

Shivam , 11 Years ago

Grade 12

1 Answers

Jitender Singh

Ans:
Hello Student,
Please find answer to your question below

Curve 1:
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Differentiate
$\frac{2x}{a^{2}} + \frac{2y}{b^{2}}.\frac{dy}{dx} = 0$
Slope of tangent:
$\frac{dy}{dx} = \frac{-b^{2}x}{a^{2}y}$ …......(1)
Curve 2:
$\frac{x^{2}}{l^{2}} - \frac{y^{2}}{m^{2}} = 1$
Differentiate
$\frac{2x}{l^{2}} - \frac{2y}{m^{2}}.\frac{dy}{dx} = 0$
Slope of tangent:
$\frac{dy}{dx} = \frac{m^{2}x}{l^{2}y}$ ...........(2)
Since curve intersect orthogonally, we have
$(1)\times (2) = -1$
$\frac{b^{2}m^{2}x^{2}}{a^{2}l^{2}y^{2}} = 1$
Since curve are intersecting, to find intersection
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{x^{2}}{l^{2}} - \frac{y^{2}}{m^{2}}$
$x^{2}(\frac{1}{a^{2}}-\frac{1}{l^{2}}) = y^{2}(-\frac{1}{b^{2}}-\frac{1}{m^{2}})$
$\frac{x^{2}}{y^{2}}(\frac{1}{a^{2}}-\frac{1}{l^{2}}) = (-\frac{1}{b^{2}}-\frac{1}{m^{2}})$
$\frac{a^{2}l^{2}}{b^{2}m^{2}}(\frac{1}{a^{2}}-\frac{1}{l^{2}}) = (-\frac{1}{b^{2}}-\frac{1}{m^{2}})$
$l^{2}-a^{2} = -m^{2}-b^{2}$
$a^{2} -b^{2} = l^{2} + m^{2}$
Option © is correct.