Equation of Pair of Tangents: Ellipse
Trending Questions
- tan−1(a+b√ab)
- tan−1(a−b2√ab)
- tan−1(a−b√ab)
- tan−1(2√ab)
- x2a2+y2b2−2xa=0
- x2a2+y2b2+2xa=0
- x2a2+y2b2+2yb=0
- x2a2+y2b2−2yb=0
- y−√2=2√2(x−π4)
- y+√2=√2(x+π4)
- y−√2=−√2(x−π4)
- y−√2=√2(x−π4)
- x+y+5=0
- x−y+5=0
- x−y−5=0
- All of the above.
x2+y2+2αx+2βy−2α−1=0. Then
- The locus of perpendicular tangents to the ellipse is (x−1)2+(y+2)2=7
- The locus of perpendicular tangents to the ellipse is (x−1)2+(y+2)2=19
- Area of auxiliary circle is 2π sq. unit
- Eccentricity of the ellipse is 12
- x+√3y=4
- x=4
- x+2√2y=6
- y=2
- L:2y=1
- a=1 units
- a=1√2 units
- L:y=1
A set of parallel chords of the parabola y2=4ax have their mid points on
any straight line through the vertex
any straight line through the focus
a straight line parallel to the axis
another parabola
- a straight line
- a parabola
- an ellipse
- a circle
Find the equation of pair of tangents to the ellipse x225+y216=1 from (5, 4).
5x + 4y + xy = 61
4x + 5y - xy = 21
4x + 5y - xy = 20
4x + 5y + xy = 40
- (4, 4)
- (−1, 2)
- None of these
- (9/4, 3/8)
Find the equation of pair of tangents to the ellipse x225+y216=1 from (5, 4).
4x + 5y - xy = 20
4x + 5y - xy = 21
4x + 5y + xy = 40
5x + 4y + xy = 61
- R=√e
- A=−83e√e
- R=2e
- A=√e6(2e−3)+1
- (2a+x)y2+4a3=0
- (x+2a)y2+4a2=0
- (y+2a)x2+4a3=0
- None of these
- x2a2+y2b2+2xa=0
- x2a2+y2b2+2yb=0
- x2a2+y2b2−2xa=0
- x2a2+y2b2−2yb=0
- (2, 3√32), (4, 0)
- (2, √3√2), (4, √32)
- (4, 3√3√3), (2, 0)
- (2, 0), (4, 0)
- 5√52 unit
- √1572 unit
- √612 unit
- √2212 unit
Find the equation of pair of tangents to the ellipse x225+y216=1 from (5, 4)
4x+5y−xy=20
4x+5y−xy=21
4x+5y+xy=40
5x+4y+xy=2
- The locus of perpendicular tangents to the ellipse is (x−1)2+(y+2)2=7
- The locus of perpendicular tangents to the ellipse is (x−1)2+(y+2)2=19
- Area of auxiliary circle is 2π sq. unit
- Eccentricity of the ellipse is 12
Find the equation of pair of tangents to the ellipse x225+y216=1 from (5, 4)
4x + 5y - xy = 20
4x + 5y + xy = 40
5x + 4y + xy = 61
4x + 5y - xy = 21