Tangents are drawn from the point P(3,4) to the ellipse x²/9 + y²/4 = 1 touching the ellipse at points A and B. The orthocentre of the ΔPAB is (5,87), (75,258), (115,85), (825,75)

In the given problem, we have to draw the chord of contact AB and find its equation, which is described by the equation T=0. Here, P(3,4) and from equation of ellipse, we write 3x/9 + 4y/4 = 1 ⇒ x/3 + y/1 = 1 ⇒ x + 3y = 3 (i) Now we have to determine the point of intersection of this chord and ellipse x²/9 + y²/4 = 1 (ii) From eq (i) and (ii) ⇒ (3 - 3y)²/9 + y²/4 = 1 ⇒ 9(1 - y)²/9 + y²/4 = 1 ⇒ (1 - y)²/1 + y²/4 = 1 ⇒ (1 + y² - 2y) + y²/4 = 1 ⇒ 5y²/4 - 2y = 0 ⇒ y(5y/4 - 2) = 0 ⇒ y = 0, 8/5 Putting this in (i), we get x = 3, 7/5 Equation of line perpendicular to AB is 3x - y = c (iii) Altitude will pass through Point P (3, 4) so, put the value of x and y from point P in Eq (iii) we get c = 5 Hence 3x - y = 5 (iv) Therefore we know Point A (3,0) and P (3,4), thus we will find equation of PA passing through these points, which will be x = 3, Equation of line perpendicular to PA and passing through B (7/5, 8/5) will be y = 8/5 (v) (iv) and (v) are altitudes of ΔPAB Thus, their point of intersection gives orthocenter : (115,85)