Question
Let E be the ellipse 
+ 
= 1 and C be the circle x2 + y2 = 9. Let P and Q be the points (1, 2) and (2, 1) respectively. Then -
- Q lies inside C but outside E
- Q lies outside both C and E
- P lies inside both C and E
- P lies inside C but outside E
Hint:
substitute the points on the curves and find out the values .
The correct answer is: P lies inside C but outside E
P lies inside C but outside E
On substituting the points P and Q on the circle equation, we get
1+4=5<9 and 4+1=5 <9.
Hence, P and Q both lie inside the circle C.
Now, for the ellipse,
4x2+9y2=36
P: 4+9(4)= 40 > 36.
Q: 4(4)+9=25<36
Hence, P lies outside ellipse and Q lies inside the ellipse E
if the points lie inside a curve, then it gives a negative value. if it lies outside , it gives a positive value.
Related Questions to study
The area of quadrilateral formed by tangents at the ends of latus-rectum of the ellipse x2 + 2y2 = 2 is-
the tangents are symmetrical about the origin. hence, the area of one of the triangles can be multiplied by 4 to get the total area.
The area of quadrilateral formed by tangents at the ends of latus-rectum of the ellipse x2 + 2y2 = 2 is-
the tangents are symmetrical about the origin. hence, the area of one of the triangles can be multiplied by 4 to get the total area.
The locus of mid-points of a focal chord of the ellipse 
= 1 is-
the equation of chord with a given midpoint is:
xx1/a2+yy1/b2=x12/a2+y12/b2
where (x1,y1) is the midpoint.
The locus of mid-points of a focal chord of the ellipse = 1 is-
the equation of chord with a given midpoint is:
xx1/a2+yy1/b2=x12/a2+y12/b2
where (x1,y1) is the midpoint.
The equation of the chord of the ellipse 2x2 + 5y2 = 20 which is bisected at the point (2,1) is-
The equation of the chord of the ellipse 2x2 + 5y2 = 20 which is bisected at the point (2,1) is-
The locus of the point of intersection of the perpendicular tangents to the ellipse + = 1 is
perpendicular tangents are drawn from the director circle
The locus of the point of intersection of the perpendicular tangents to the ellipse + = 1 is
perpendicular tangents are drawn from the director circle
From the point (
, 3) tangents are drawn to + = 1 and are perpendicular to each other then =
perpendicular tangents are drawn from the director circle
From the point (, 3) tangents are drawn to + = 1 and are perpendicular to each other then =
perpendicular tangents are drawn from the director circle
Two perpendicular tangents drawn to the ellipse
+ 
= 1 intersect on the curve -
the tangents drawn from the director circle intersect at right angle.
Two perpendicular tangents drawn to the ellipse + = 1 intersect on the curve -
the tangents drawn from the director circle intersect at right angle.
The angle between the pair of tangents drawn to the ellipse 3x2+ 2y2 = 5 from the point (1,2) is-
use the equation of S to find the value of theta.
The angle between the pair of tangents drawn to the ellipse 3x2+ 2y2 = 5 from the point (1,2) is-
use the equation of S to find the value of theta.
Tangents are drawn from a point on the circle x2 + y2 = 25 to the ellipse 9x2 + 16y2 – 144= 0 then find the angle between the tangents.
tangents drawn from director circle to the ellipse make an angle of 90 degrees with themselves.
Tangents are drawn from a point on the circle x2 + y2 = 25 to the ellipse 9x2 + 16y2 – 144= 0 then find the angle between the tangents.
tangents drawn from director circle to the ellipse make an angle of 90 degrees with themselves.
The equation of the normal to the ellipse 
+ 
= 1, at the point (a cosθ, b sinθ) is-
differentiating a curve gives s the slope of tangent from where the slope of normal can be found out.
The equation of the normal to the ellipse + = 1, at the point (a cosθ, b sinθ) is-
differentiating a curve gives s the slope of tangent from where the slope of normal can be found out.
The equation of the normal at the point (2, 3) on the ellipse 9x2 + 16y2 = 180 is-
differentiating a curve gives us the slope of the tangent at any point in the curve. from that, we can find the slope of normal which will help s find the equation of normal at any point on the curve.
The equation of the normal at the point (2, 3) on the ellipse 9x2 + 16y2 = 180 is-
differentiating a curve gives us the slope of the tangent at any point in the curve. from that, we can find the slope of normal which will help s find the equation of normal at any point on the curve.
The equation of tangent to the ellipse x2 + 3y2 = 3 which is 
r to line 4y = x – 5 is-
slope of parallel lines are equal and product of perpendicular lines is -1.
this can be used to find the required slope.
The equation of tangent to the ellipse x2 + 3y2 = 3 which is r to line 4y = x – 5 is-
slope of parallel lines are equal and product of perpendicular lines is -1.
this can be used to find the required slope.
The equation of the tangents to the ellipse 4x2 + 3y2 = 5 which are parallel to the line y = 3x + 7 are
parallel lines have exactly equal slopes and perpendicular lines have the product of their slopes =-1. from this, we can find out the slope of the tangent of the ellipse.
The equation of the tangents to the ellipse 4x2 + 3y2 = 5 which are parallel to the line y = 3x + 7 are
parallel lines have exactly equal slopes and perpendicular lines have the product of their slopes =-1. from this, we can find out the slope of the tangent of the ellipse.
The ellipse + = 1 and the straight line y = mx + c intersect in real points only if-
since the curves intersect at real points only, D should be greater than 0
The ellipse + = 1 and the straight line y = mx + c intersect in real points only if-
since the curves intersect at real points only, D should be greater than 0
The line x cos 
+ y sin = p will be a tangent to the conic + = 1, if-
the tangent touches the curve at a single point and hence, has a single real solution to the equation.
The line x cos + y sin = p will be a tangent to the conic + = 1, if-
the tangent touches the curve at a single point and hence, has a single real solution to the equation.
Find the equations of tangents to the ellipse 9x2 + 16y2 = 144 which pass through the point (2,3).
equation of tangent of an ellipse is given by :
y= m+ √(a2m2+b2)
this is used to find the values of m.
Find the equations of tangents to the ellipse 9x2 + 16y2 = 144 which pass through the point (2,3).
equation of tangent of an ellipse is given by :
y= m+ √(a2m2+b2)
this is used to find the values of m.
