Slope Form of Tangent: Hyperbola
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What is the condition for a line y=mx+c to be tangent to the hyperbola x2a2−y2b2=1.
c=±√b2−a2m2
c=±√a2m2−b2
c=±√a2−b2m2
c=±√b2+a2m2
- a, 4, 1
- a, 4, 2
- 2a, 8, 1
- 2a, 4, 1
- 1
- 2
- −1
- −12
What is the equation of tangent to the hyperbola
2x2 − 3y2 = 6 which is parallel to the line y = 3x + 4, is
y = 3x − 5
y = 3x + 5 and y = 3x − 5
- None of the above
y = 3x + 5
Column−1Column−2Column−3(I)x2+y2=a2(i)my=m2x+a(P)(am2, 2am)(II)x2+a2y2=a2(ii)y=mx+a√m2+1(Q)(−ma√m2+1, a√m2+1)(III)y2=4ax(iii)y=mx+√a2m2−1(R)(−a2m√a2m2+1, 1√a2m2+1)(IV)x2−a2y2=a2(iv)y=mx+√a2m2+1(S)(−a2m√a2m2−1, −1√a2m2−1)
For a=√2, if a tangent is drawn to a suitable conic (Column 1) at the point of contact (−1, 1), then which of the following options is the only CORRECT combination for obtaining its equation?
- (I)(i)(P)
- (I)(ii)(Q)
- (II)(ii)(Q)
- (III)(i)(P)
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
- (x2a2−y2b2)=(x2a2+y2b2)2
- 2(x2a2−y2b2)=(x2a2+y2b2)2
- x2a2+y2b2=(x2a2−y2b2)2
- 2x2a2−y2b2=(x2a2+y2b2)2
What is the equation of tangent to the hyperbola
2x2 − 3y2 = 6 which is parallel to the line y = 3x + 4, is
y = 3x + 5
y = 3x − 5
y = 3x + 5 and y = 3x − 5
- None of the above
- x+y+3√3 = 0
- x+y+3√2 = 0
- x+y−3√3 = 0
- x+y−3√2 = 0
- y=±3x√7+±15√7
- y=±3√27x±15√7
- y=±2√27x±15√7
- y=±4x√7±15√7
- a, 4, 1
- a, 4, 2
- 2a, 8, 1
- 2a, 4, 1
- Quadrilateral formed by the points of contact of the tangent is rectangle.
- Slope of one pair of parallel tangent is 52
- Slope of one pair of parallel tangent is −52
- area formed by the points of contact of the tangent is 180 sq. units
The equations to the common tangents to the two hyperbolas x2a2−y2b2=1 and y2a2−x2b2=1 are
y=±x±√(b2−a2)
y=±x±√(a2−b2)
y=±x±(a2−b2)
y=±x±√(a2+b2)
The line xcosα+ysinα=p touches the hyperbola
x2a2−y2b2=1 if
a2cos2α−b2sin2α=p2
- a2cos2α−b2sin2α=p
- a2cos2α−b2cos2α=2p
- a2cos2α+b2sin2α=p
- 0
- 1
- 2
- 3