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The lines y = -3/2 x and y = - 2/5x intersect the curve 3x 2 + 4xy + 5y 2 – 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

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
  1. Intersect each other at angle of 45°
  2. Are parallel to each other
  3. Are perpendicular to each other
  4. None of these

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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 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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