Point Form of Normal:Hyperbola
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Let P(6, 3) be a point on the hyperbola x2a2−y2b2=1. If the normal at the point P intersects the X-axis at (9, 0), then the eccentricity of the hyperbola is
√52
√32
√2
√3
- √72
- √32
- √5
- √3
- 3x−4y=4
- 3x+4y=8
- 5x−4y=9
- 6x−3y=4
Let P(6, 3) be a point on the hyperbola x2a2−y2b2=1.
If the normal at the point P intersects the x-axis at (9, 0), then the eccentricity of hyperbola is,
√52
√32
√2
√3
Ahyperbola x225−y216=1 is given and a normal is drawn at the point (5√3, 4√2).
What is the abscissa of the point at which it meets the x-axis.
2899
191125√3
156√35
3150√3
- 9x + 4y = 40
- 9x - 4y = 40
- 4x + 9y = 40
- 4x - 9y = 40
Let P(6, 3) be a point on the hyperbola x2a2−y2b2=1.
If the normal at the point P intersects the x-axis at (9, 0), then the eccentricity of hyperbola is,
√52
√32
√2
√3
Without using Pythagoras theorem, show that , and are the vertices of a right angled triangle.