Question
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.
Hint:
find the equation director circle of the ellipse.
The correct answer is:
π/2
Equation of circle: x2+y2=25
Equation of ellipse : 9x2+16y2-144=0
a = 4 and b = 3
equation of director circle : x2+y2= a2+b2
x2+y2= 42+32 =25 = equation of given circle.
Therefore, the given circle is the director circle of the given ellipse.
We know that tangents drawn from director circle to the ellipse make an angle of 90 degrees with themselves. Hence , answer is π/2
tangents drawn from director circle to the ellipse make an angle of 90 degrees with themselves.
Related Questions to study
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.
Find the equation of the tangent to the ellipse x2 + 2y2 = 4 at the points where ordinate is 1.
the standard form of equation of tangents is obtained by substituting the values of coordinates into the equation of the curve
Find the equation of the tangent to the ellipse x2 + 2y2 = 4 at the points where ordinate is 1.
the standard form of equation of tangents is obtained by substituting the values of coordinates into the equation of the curve
If 
+ 
= 
touches the ellipse + = 1, then its eccentric angle θ is equal to-
the standard equation of an ellipse is obtained by substituting the parametric point on the curve equation.
If + = touches the ellipse + = 1, then its eccentric angle θ is equal to-
the standard equation of an ellipse is obtained by substituting the parametric point on the curve equation.
The position of the point (4,– 3) with respect to the ellipse 2x2 + 5y2 = 20 is-
the position of a point with respect to a curve can be calculated by substituting the point and finding the resultant value
if it is less than zero, then it lies inside the curve
equal to zero then lies on the curve
greater than zero then lies outside the curve
The position of the point (4,– 3) with respect to the ellipse 2x2 + 5y2 = 20 is-
the position of a point with respect to a curve can be calculated by substituting the point and finding the resultant value
if it is less than zero, then it lies inside the curve
equal to zero then lies on the curve
greater than zero then lies outside the curve
The parametric representation of a point on the ellipse whose foci are (– 1, 0) and (7,0) and eccentricity 1/2 is-
the parametric form of a point on the ellipse gives us the coordinates of any point on the ellipse for a given angle.
The parametric representation of a point on the ellipse whose foci are (– 1, 0) and (7,0) and eccentricity 1/2 is-
the parametric form of a point on the ellipse gives us the coordinates of any point on the ellipse for a given angle.
Let P be a variable point on the ellipse 
+ 
=1 with foci S and S'. If A be the area of triangle pss', then maximum value of A is–
the maxima or minima of a function is calculated by finding out the critical points of the function and then substituting the value of the critical point in the function.
Let P be a variable point on the ellipse + =1 with foci S and S'. If A be the area of triangle pss', then maximum value of A is–
the maxima or minima of a function is calculated by finding out the critical points of the function and then substituting the value of the critical point in the function.
If S and S' are two foci of an ellipse + = 1 (a < b) and P (x1, y1) a point on it, then SP + S' P is equal to-
the length of major axis of an ellipse is 2a for a horizontal ellipse and 2b for a vertical ellipse. a horizontal ellipse is formed when a>b and a vertical ellipse is formed when b>a. the major axis lies along x axis when a>b and along y axis when b>a.
If S and S' are two foci of an ellipse + = 1 (a < b) and P (x1, y1) a point on it, then SP + S' P is equal to-
the length of major axis of an ellipse is 2a for a horizontal ellipse and 2b for a vertical ellipse. a horizontal ellipse is formed when a>b and a vertical ellipse is formed when b>a. the major axis lies along x axis when a>b and along y axis when b>a.
The eccentricity of the ellipse represented by the equation 25x2 + 16y2 – 150x – 175 = 0 is -
the given equation has to be converted into the whole square form for x and y so that the standard form of ellipse can be observed. from the standard form, the eccentricity can be calculated by the values of a and b.
The eccentricity of the ellipse represented by the equation 25x2 + 16y2 – 150x – 175 = 0 is -
the given equation has to be converted into the whole square form for x and y so that the standard form of ellipse can be observed. from the standard form, the eccentricity can be calculated by the values of a and b.
The foci of the ellipse,25 (x + 1)2 + 9 (y + 2)2 = 225, are at-
the ellipse has its vertex at (-1,-2). The terms x+1 and y+2 can be replaced with X and Y for better understanding so that it gets converted into the standard form.
The foci of the ellipse,25 (x + 1)2 + 9 (y + 2)2 = 225, are at-
the ellipse has its vertex at (-1,-2). The terms x+1 and y+2 can be replaced with X and Y for better understanding so that it gets converted into the standard form.
