The Length Of The Common Tangent To The Ellipse X 225 Y 24 1 And

The Length Of The Common Tangent To The Ellipse X 225 Y 24 1 And The common tangent to the circles x 2 y 2 = 4 and x 2 y 2 6 x 8 y − 24 = 0 intersects the coordinate axes at a and b respectively. if o a and o b are equal to half of the length of the major and minor axes of an ellipse respectively, where o is the origin, then the eccentricity of the ellipse is. Algebra. graph (x^2) 25 (y^2) 16=1. x2 25 y2 16 = 1 x 2 25 y 2 16 = 1. simplify each term in the equation in order to set the right side equal to 1 1. the standard form of an ellipse or hyperbola requires the right side of the equation be 1 1. x2 25 y2 16 = 1 x 2 25 y 2 16 = 1. this is the form of an ellipse.

If L Is The Length Of The Intercept Made By A Common Tangent To The Slope form of a tangent to an ellipse; if the line y = mx c touches the ellipse x 2 a 2 y 2 b 2 = 1, then c 2 = a 2 m 2 b 2. the straight line y = mx ∓ √[a 2 m 2 b 2] represents the tangents to the ellipse. point form of a tangent to an ellipse; the equation of the tangent to an ellipse x 2 a 2 y 2 b 2 = 1 at the point (x. Equation of the ellipse is 25 x 2 9 y 2 = 1 any point on the ellipse is (5 cos θ, 3 s in ⁡ θ) at which the equation of the tangent is 5 x cos θ 3 y s in ⁡ θ = 1 if it is also a tangent to the circle x 2 y 2 = 16, then its distance from the origin is equal to the radius, ⇒ (25 co s 2 θ 9 s i n 2 θ ) 1 = 4 ⇒ 25 co s 2 θ 9. Any tangent to ellipse x 2 25 y 2 4 = 1 with slope m is : y = m x ± √ 25 m 2 4 ⋯ (1) if this is also a tangent to the circle x 2 y 2 = 16, then distance of tangent from centre of cricle = 4 ⇒ ∣ ∣ ∣ 0 − m × 0 ± √ 25 m 2 4 √ 1 m 2 ∣ ∣ ∣ = 4 ⇒ m = ± 2 √ 3 since common tangent is in the first quadrant, so. The equation is an ellipse centered at the origin. the slope at each point is x 4y. the slope is undefined when y=0, i.e. when x=±4. this is where the ellipse crosses the x axis and the tangent is, in fact, perpendicular to the x axis.

If L Is The Length Of The Intercept Made By A Common Tangent To The Any tangent to ellipse x 2 25 y 2 4 = 1 with slope m is : y = m x ± √ 25 m 2 4 ⋯ (1) if this is also a tangent to the circle x 2 y 2 = 16, then distance of tangent from centre of cricle = 4 ⇒ ∣ ∣ ∣ 0 − m × 0 ± √ 25 m 2 4 √ 1 m 2 ∣ ∣ ∣ = 4 ⇒ m = ± 2 √ 3 since common tangent is in the first quadrant, so. The equation is an ellipse centered at the origin. the slope at each point is x 4y. the slope is undefined when y=0, i.e. when x=±4. this is where the ellipse crosses the x axis and the tangent is, in fact, perpendicular to the x axis. Your approach would have been correct if the point lied outside the ellipse but since it lies inside the ellipse your method only calculates the square of one part of the chord on which the point lies because since an ellipse is symmetrical you can draw two equal chords through a point.to solve this correctly remember that the diameter of a ellipse through a point is largest among all chords. We know that the vertices and foci are related by the equation c2 = a2 − b2. solving for b2, we have: c2 = a2 − b2 25 = 64 − b2 substitute for c2 and a2 b2 = 39 solve for b2. now we need only substitute a2 = 64 and b2 = 39 into the standard form of the equation. the equation of the ellipse is x2 64 y2 39 = 1.