Slope Form of Tangent: Hyperbola
Trending Questions
Q. The locus of the foot of perpendicular from the centre upon any normal to the hyperbola x2a2−y2b2=1 is
- (x2−y2)2(a2y2−b2x2)=(a2+b2)2x2y2
- (x2+y2)2(a2y2−b2x2)=(a2+b2)2x2y2
- (x2−y2)2(a2y2−b2x2)=(a2−b2)2x2y2
- (x2+y2)2(a2y2−b2x2)=(a2−b2)2x2y2
Q.
The equations of the common tangents to the two hyperbolas
x2a2−y2b2=1(a<b) and y2a2−x2b2=1 (a>b) is
- y=±x±√a2+b2
- y=±x±√a3+b3
- y=±x±√a2−b2
- y=±x±√a3−b3
Q. A tangent to the hyperbola x2a2−y2b2=1 cuts the ellipse x2a2+y2b2=1 in points P & Q. The locus of the mid- point PQ is
- (x2a2−y2b2)=(x2a2+y2b2)2
- 2(x2a2−y2b2)=(x2a2+y2b2)2
- x2a2+y2b2=(x2a2−y2b2)2
- 2x2a2−y2b2=(x2a2+y2b2)2
Q.
The circle x2+y2−8x=0 and hyperbola x29−y24=1 intersect at the points A and B.
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
2x−5√y−20=0
2x−√5y+4=0
3x−4y+8=0
4x−3y+4=0
Q.
If is a tangent to both the parabolas, and , then is equal to
Q. The angle between lines joining the origin to the points of intersection of the line √3x+y=2 and the curve y2−x2=4 is
- tan−1(√32)
- π2
- tan−1(2√3)
- π6
Q. The total number of tangents to the hyperbola x29−y24=1 that are perpendicular to the line 5x+2y−3=0 is
- 2
- 1
- 3
- 0
Q. The equation of a tangent to the hyperbola 4x2−5y2=20 parallel to the line x−y=2 is :
- x−y+1=0
- x−y+7=0
- x−y−3=0
- x−y+9=0
Q. The slope(s) of the common tangent(s) to the two hyperbolas x2a2−y2b2=1 and y2a2−x2b2=1 is/are
- 1
- 2
- −1
- −12
Q.
If in a , the altitudes from the vertices on the opposite side are in HP, then and are in
HP
Arithmetic – Geometric Progression
AP
GP
Q.
The points on the curve where the tangent is vertical (parallel to theaxis), are
Q. The equation of the common tangent to x2=6y and 2x2−4y2=9 can be
- x+y=1
- x−y=1
- x−y=32
- x+y=92
Q. If P and Q be two points on the hyperbola x2a2−y2b2=1, whose centre is C such that CP is perpnediuclar to CQ, a<b, then the value of 1CP2+1CQ2 is
- b2−a22ab
- 1a2+1b2
- 1a2−1b2
- 2abb2−a2
Q. The equations of the tangents to the hyperbola x2−4y2=36 which are perpendicular to the line x−y+4=0 are:
- x+y+3√3 = 0
- x+y+3√2 = 0
- x+y−3√3 = 0
- x+y−3√2 = 0