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Integral Calculus> The lines y = -3/2 x and y = - 2/5x inter...

question mark

The lines y = -3/2 x and y = - 2/5x intersect the curve 3x² + 4xy + 5y² – 4 = 0 at the points P and Q respectively. The tangents drawn to the curve at P and Q:

options

Intersect each other at angle of 45°

Are parallel to each other

Are perpendicular to each other

None of these

Aditya Kartikeya , 10 Years ago

Grade 10

1 Answers

Jitender Singh

Last Activity: 10 Years ago

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

$3x^2+4xy+5y^2-4=0$
Differentiate
$6x+4xy'+4y+10yy'=0$
$y'(4x+10y) = -(6x+4y)$
$y' = \frac{-(6x+4y)}{4x+10y}$
Slope of the tangent at intersection by
$y = -\frac{3}{2}x$
$y' = \frac{-(6x+4(-\frac{3x}{2}))}{4x+10(-\frac{3x}{2})}$
$y' = 0$
Parallel to x-axis.
Slope of the tangent at intersection by
$y = \frac{-2x}{5}$
$y' = \frac{-(6x+4(-\frac{2x}{5}))}{4x+10(\frac{-2x}{5})}$
$y' = \frac{-(6x+4(-\frac{3x}{2}))}{0}$
Parallel to y-axis.
So both tangent are perpendicular.
Option (3) is correct.

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