Slope Form of Tangent: Hyperbola
Trending Questions
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. Tangents are drawn from a point P to the parabola y2=4ax. If the chord of contact of the parabola is a tangent to the hyperbola x2a2−y2b2=1, then the locus of P is
- a2x2=a4−y2b2
- 4a2x2=4a4−y2b2
- 4a2x2=a4−y2b2
- a2x2=4a4−y2b2
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 equation of tangents to the curve 9y2−4x2=1 which is parallel to the line 3y=x+7 is/are
- 2x−6y=√3
- 2x−6y=−√3
- 16x−20y=25
- 16x−20y=−25
Q. The equation of the common tangent to x2=6y and 2x2−4y2=9 can be
- x+y=1
- x−y=1
- x+y=92
- x−y=32
Q. For all real permissible values of m, if the straight line y=mx+√9m2−4 is tangent to a hyperbola, then equation of the hyperbola can be
- 9x2−4y2=64
- 4x2−9y2=64
- 9x2−4y2=36
- 4x2−9y2=36
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(2√3)
- π6
- tan−1(√32)
- π2
Q. If the distance between two parallel tangents drawn to the hyperbola x29−y249=1 is 2 unit, then which of the following is(are) correct?
- 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
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
- 2abb2−a2
- 1a2−1b2
Q. If the line 25x+12y−45=0 meets the hyperbola 25x2−9y2=225 only at point (a, −5λ3), then the value of a+λ is
Q. The circle x2+y2−8x=0 and hyperbola x29−y24=1 intersects at the points A and B. Then
- equation of common tangents will be 2x±√5y+4=0
- point of intersection of common tangents is (−2, 0)
- equation of common tangents will be 2x±√5y−4=0
- intersection point of common tangents will be (2, 0)
Q. Equation of tangent(s) which passing through (2, 8) to the hyperbola 5x2−y2=5 is (are)
- y=3x−2
- 23x−3y−22=0
- 23x−3y+22=0
- y=3x+2
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. 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 equation of tangents to the curve 9y2−4x2=1 which is parallel to the line 3y=x+7 is/are
- 2x−6y=√3
- 2x−6y=−√3
- 16x−20y=25
- 16x−20y=−25
Q. Let x1=11⋅3+13⋅5+15⋅7+⋯ upto 10 terms,
y1=Sum of roots of equation of x2−7x+10=0. If (x1, y1) lies inside the hyperbola (21x)2100−y2(7b)2=−1, then number of integeral values for b is
y1=Sum of roots of equation of x2−7x+10=0. If (x1, y1) lies inside the hyperbola (21x)2100−y2(7b)2=−1, then number of integeral values for b is
Q. If tangents are drawn to the hyperbola 4x2−5y2=20 parallel to the line x−y=2, then
- equation of tangent is x−y=1
- point of contact will be (−5, −4)
- point of contact will be (5, 4)
- equation of tangent is x−y+1=0
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. If the value's of m for which the line y=mx+2√5 touches the hyperbola 16x2−9y2=144 are roots of the equation x2−(a+b)x−4=0, then the value of a+b=
Q. If 2x−y+1=0 is a tangent to the hyperbola x2a2−y216=1, then which of the following CANNOT be sides of a right angled triangle?
- a, 4, 1
- a, 4, 2
- 2a, 8, 1
- 2a, 4, 1
Q. Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.
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?
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)