Write the parametric equation of ellipse x - 3 2 25 + y + 2 2 16 = 1 .

For ellipse x - h 2 a 2 + y - k 2 b 2 = 1 . Parametric Equation will be x = h + a cos θ , y = k + b sin θ . Therefore, For Ellipse x - 3 2 25 + y + 2 2 16 = 1 , a = 5 , b = 4 and h , k = 3 , - 2 Parametric Equation will be, x = 3 + 5 cos θ and y = - 2 + 4 cos θ

Write the parametric equation of ellipse x - 3 2 25 + y + 2 2 16 = 1 .

For ellipse x - h 2 a 2 + y - k 2 b 2 = 1 . Parametric Equation will be x = h + a cos θ , y = k + b sin θ . Therefore, For Ellipse x - 3 2 25 + y + 2 2 16 = 1 , a = 5 , b = 4 and h , k = 3 , - 2 Parametric Equation will be, x = 3 + 5 cos θ and y = - 2 + 4 cos θ

Prove that the distance of the point 6 cos θ , 2 sin θ on the ellipse x 2 6 + y 2 2 = 1 from the centre of the ellipse is 2 , if θ = 5 π 4 .

Given ellipse x 2 6 + y 2 2 = 1 Also given the parametric form of co-ordinates of point P As P : 6 cos θ , 2 sin θ and O : 0 , 0 and provided θ = 5 π 4 O P = 6 cos θ 2 + 2 sin θ 2 O P = 6 cos 2 θ + 2 sin 2 θ O P = 6 cos θ 2 + 2 sin θ 2 O P = 6 - 1 2 2 + 2 - 1 2 2 ⇒ O P = 6 × 1 2 + 2 × 1 2 O P = 4 O P = 2 Hence, proved.

Find the locus of the middle points of chords of an ellipse x 2 a 2 + y 2 b 2 = 1 a > b , which are drawn through the positive end of the minor axis.

Let a cos ϕ , b sin ϕ be the co-ordinates of the other extremity of the chord of ellipse x 2 a 2 + y 2 b 2 = 1 a > b . The positive end of the minor axis is clearly 0 , b . Let ( x , y ) be the middle point of chord, then x = a cos ϕ + 0 2 and y = b + b sin ϕ 2 Hence, cos ⁡ ϕ = 2 x a … 1 and sin ⁡ ϕ = 2 y - b b … 2 Squaring and adding equations ( 1 ) and ( 2 ) , we get 1 = 4 x 2 a 2 + 2 y - b b 2 Or 4 x 2 a 2 + 4 y 2 - 4 y b + b 2 b 2 - 1 = 0 Simplifying, we get the required locus of ( x , y ) as x 2 a 2 + y 2 b 2 = y b . This is the locus of the middle points of chords of an ellipse which are drawn through the positive end of minor axis ( 0 , b ) .

Check whether the line 4 x + 5 y = 40 touches the ellipse x 2 50 + y 2 32 = 1 or not. If yes, then also find its point of contact.

Equation of ellipse S : x 2 50 + y 2 32 = 1 ⇒ a 2 = 50 , b 2 = 32 . Equation of tangent given is 4 x + 5 y = 40 ⇒ y = - 4 x 5 + 40 5 … 1 Equation of tangent to an ellipse in slope form is y = m x ± a 2 m 2 + b 2 . For equation 1 to be tangent they must be similar a 2 m 2 + b 2 = 40 5 = 8 and m = - 4 5 ⇒ LHS = a 2 16 25 + b 2 = 50 × 16 25 + 32 = 8 = RHS Therefore, we can say the line 4 x + 5 y = 40 touches the ellipse. Let the point of contact of tangent and ellipse is h , k then equation in point form is h x 50 + k y 32 = 1 comparing it with 4 x + 5 y = 40 ⇒ x 10 + y 8 = 1 we get, h 50 = 1 10 ⇒ h = 5 and k 32 = 1 8 ⇒ k = 4 Therefore, point of contact of tangent and ellipse is 5 , 4 .