Any chord of the conic `x^(2)+y^(2)+xy=1` passing through origin is bisected at a point (p, q), then `(p+q+12)` equals to :
A
13
B
14
C
11
D
12
Text Solution
Verified by Experts
The correct Answer is:
D
Topper's Solved these Questions
ELLIPSE
VK JAISWAL|Exercise Exercise-2 : Comprehension Type Problems|5 Videos
ELLIPSE
VK JAISWAL|Exercise Exercise-3 : Matching Type Problems|1 Videos
DIFFERENTIAL EQUATIONS
VK JAISWAL|Exercise EXERCISE (SUBJECTIVE TYPE PROBLEMS)|6 Videos
FUNCTION
VK JAISWAL|Exercise SUBJECTIVE TYPE PROBLEMS|34 Videos
Similar Questions
Explore conceptually related problems
Statement 1: Any chord of the conic x^(2)+y^(2)+xy=1 through (0,0) is bisected at (0,0) .Statement 2: The center of a conic is a point through which every chord is bisected.
The circumference of the circle x^(2)+y^(2)-2x+8y-q=0 is bisected by the circle x^(2)+y^(2)+4x+12y+p=0, then p+q is equal to
A point P moves such that the chord of contact of P with respect to the circle x^(2)+y^(2)=4 passes through the point (1, 1). The coordinates of P when it is nearest to the origin are
Tangent is drawn at any point (p,q) on the parabola y^(2)=4ax. Tangents are drawn from any point on this tangant to the circle x^(2)+y^(2)=a^(2), such that the chords of contact pass through a fixed point (r,s). Then p,q,r ands through a fixed point (r,s). Then p,q,rrq^(2)=4ps^(2)( C) rq^(2)=-4ps^(2)(D)r^(2)q=-4p^(2)s
The middle point of chord x+3y=2 of the conic x^(2)+xy-y^(2)=1, is
PQ is a chord of the circle x^(2)+y^(2)-2x-8=0 whose mid-point is (2, 2). The circle passing through P, Q and (1, 2) is
Find the equation of a curve passing through the point (1.1) if the perpendicular distance of the origin from the normal at any point P(x,y) of the curve is equal to the distance of P from the x-axis.
The graph of the conic x^(2)-(y-1)^(2)=1 has one tangent line with positive slope that passes through the origin. The point of tangency being (a, b). Q. Length of the latusrectum of the conic is
VK JAISWAL-ELLIPSE -Exercise-4 : Subjective Type Problems
