TANGENTS AND NORMALS show that the curves x^2 / (a^2 + (lambda)1) + y^2 / (b^2 + (lambda)1) =1 and x^2 / (a^2 + (lambda)2) + y^2 / (b^2 + (lambda)2) = 1 intersect at right angles

Ans:Hello Student,Please find answer to your question belowSlope of tangent:.................(1)Slope of tangent:..............(2)Just manipulate the curve equations from here for x2/y2.

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Differential Calculus> TANGENTS AND NORMALS show that the curves...

TANGENTS AND NORMALS

show that the curves x^2 / (a^2 + (lambda)₁) + y^2 / (b^2 + (lambda)₁) =1 and x^2 / (a^2 + (lambda)₂) + y^2 / (b^2 + (lambda)₂) = 1 intersect at right angles

taniska , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

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

$\frac{x^2}{a^2+\lambda _{1}} + \frac{y^2}{b^2+\lambda _{1}} = 1$
Slope of tangent:
$\frac{2x}{a^2+\lambda _{1}} + \frac{2yy'_{1}}{b^2+\lambda _{1}} = 0$
$y'_{1} = \frac{-x(b^2+\lambda _{1})}{y(a^2+\lambda _{1})}$ .................(1)
$\frac{x^2}{a^2 + \lambda _{2}} + \frac{y^2}{b^2+\lambda _{2}} = 1$
Slope of tangent:
$\frac{2x}{a^2+\lambda _{2}} + \frac{2yy'_{2}}{b^2+\lambda _{2}} = 0$
$y'_{2} = \frac{-x(b^2+\lambda _{2})}{y(a^2+\lambda _{2})}$ ..............(2)
$(1)\times (2) = -1$
$y_{1}'y_{2}' = \frac{x^2(b^2+\lambda _{1})(b^2+\lambda _{2})}{y^2(a^2+\lambda _{2})(a^2+\lambda _{2})}$
Just manipulate the curve equations from here for x²/y^2.