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 .

Find the equation of the tangents at the ends of the latus rectum of the ellipse x 2 a 2 + y 2 b 2 = 1 and also show that they pass through the points of intersection of the major axis and directrices.

Equation of ellipse is S : x 2 a 2 + y 2 b 2 = 1 . End points of latus rectum is L a e , b 2 a and L ' a e , - b 2 a . Equation of tangent is T = 0 . Through L in point form is a e x a 2 + b 2 y a b 2 = 1 ⇒ e x + y = a . Through L ' in point form is a e x a 2 - b 2 y a b 2 = 1 ⇒ e x - y = a . The point of intersection of both the tangents is a e , 0 which is the point of intersection of major axis and directrix of ellipse.

If ' P ' be a moving point on the ellipse x 2 25 + y 2 16 = 1 in such a way that tangent at ' P ' intersect x = 25 3 at Q then circle on P Q as diameter passes through a fixed point. Find that fixed point.

Equation of ellipse is x 2 25 + y 2 16 = 1 , ⇒ a 2 = 25 , b 2 = 16 Therefore, b 2 = a 2 1 - e 2 ⇒ 16 = 25 1 - e 2 ⇒ e 2 = 1 - 16 25 ⇒ e = 3 5 Therefore, Focus of ellipse is S a e , 0 = 3 , 0 and S - a e , 0 = - 3 , 0 . Equation of directrix is x = ± a e = ± 25 3 . Therefore, the line at which Q lies on directrix of the ellipse. Let P 5 cos θ , 4 sin θ By definition of ellipse P S = e P Q . P S and P Q are perpendicular to each other. And P Q is diameter of circle therefore circle passes through focus of ellipse. Therefore, the fixed point is 3 , 0 .

Find the equation of tangents to the ellipse x 2 50 + y 2 32 = 1 , which passes through a point ( 15 , - 4 ) .

Given the equation of ellipse x 2 50 + y 2 32 = 1 . The equation of tangent passing through 15 , - 4 . From ellipse, we have a 2 = 50 , b 2 = 32 . We know the equation of tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 is y = m x ± a 2 m 2 + b 2 . . . 1 So, y - m x = ± a 2 m 2 + b 2 ( passing through 15 , - 4 ) - 4 - m × 15 = ± 50 m 2 + 32 Squaring both sides, 16 + 225 m 2 + 120 m = 50 m 2 + 32 175 m 2 + 120 m - 16 = 0 = m + 4 5 m - 4 35 So, m = - 4 5 , 4 35 The equation of the tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 is y = m x ± a 2 m 2 + b 2 Case 1 for m = - 4 5 y = - 4 5 x + 50 × 16 25 + 32 ⇒ 5 y + 4 x = 1600 ⇒ 4 x + 5 y = 40 Case 2 for m = 4 35 y = 4 35 x + 4 35 2 × 50 + 32 ⇒ 35 y = 4 x + 4000 ⇒ 4 x - 35 y = 200

If 3 x + 4 y = 12 intersect the ellipse x 2 25 + y 2 16 = 1 at P and Q , then find the point of intersection of tangents at P and Q .

Given equation of ellipse x 2 25 + y 2 16 = 1 . . . 1 So, a 2 = 25 & b 2 = 16 . And also equation of line 3 x + 4 y = 1 . . . 1 Let h , k are the point of intersection of tangent. Then equation of chord of contact h x 25 + k y 16 = 1 . . . 2 ∴ Equations 1 & 2 represent the same line. So, 3 h 25 = 4 k 16 = 12 1 So, 3 = 12 × h 25 ⇒ h = 3 × 25 12 ⇒ h = 25 4 And k 16 × 12 = 4 ⇒ k = 4 × 16 12 ⇒ k = 16 3 So, the point is 25 4 , 16 3 .