Point Form of Normal: Ellipse
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Q. On the ellipse x28+y24=1, let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x+2y=0. Let S and S′ be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS′, then the value of (5−e2)⋅A is
- 12
- 14
- 6
- 24
Q. The equation x22−r+y2r−5+1=0 represent an ellipse if
Q. The locus of mid-points of the line segments joining (−3, −5) and the points on the ellipse x24+y29=1 is
- 36x2+16y2+90x+56y+145=0
- 9x2+4y2+18x+8y+145=0
- 36x2+16y2+72x+32y+145=0
- 36x2+16y2+108x+80y+145=0
Q.
On the portion of the straight line, intercepted between the axes, a square is constructed on the side of the line away from the origin. Then the point of intersection of its diagonals has co-ordinates
Q. The equation of the normal to the ellipse x218+y28=1 at the point (3, 2) is .
- 3x - 2y = 5
- 3x + 2y = 5
- 2x - 3y = 5
- 2x + 3y = 5