Question
The equation of the normal at the point (2, 3) on the ellipse 9x2 + 16y2 = 180 is-
- 3y = 8x –10
- 3y – 8x + 7 = 0
- 8y + 3x + 7 = 0
- 3x + 2y + 7 = 0
Hint:
find out the slope of the tangent and the normal. use the slope of normal to find the equation of normal at the given point.
The correct answer is: 3y – 8x + 7 = 0
3y – 8x + 7 = 0
slope of tangent = dy/dx => 18x + 32y dy/dx = 0
dy/dx = -9x/16y
slope of normal = 16y/9x
at the point 2,3 slope = 8/3
equation of normal at 2,3 :
y-3 = 8/3(x-2)
3y-9=8x-16
3y=8x-7
differentiating a curve gives us the slope of the tangent at any point in the curve. from that, we can find the slope of normal which will help s find the equation of normal at any point on the curve.
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