Tangents are drawn from the point P(3, 4) to the ellipse touching the ellipse at points A and B.

Linked Question 1

The equation of the locus of the point whose distances from the point P and the line AB are equal, is

9x² + 9y² – 6xy – 54x – 62y – 241 = 0

x² + y² – 2xy + 27x + 31y – 120 = 0

x² + 9y² + 6xy – 54x + 62y – 241 = 0

9x² + y² – 6xy – 54x – 62y + 241 = 0

Straight line AB :

$y - 0 = \frac{- 1}{3} (x - 3)$

Applying x + 3y = 3

Applying PS = SM

${(h - 3)}^{2} + {(k - 4)}^{2} = {| \frac{h + 3 k - 3}{\sqrt{10}} |}^{2}$

$∴$ Locus of S(h, k) is

10(x² + y² – 6x – 5y + 25) = x² + 9y² + 6xy + 9 – 6(x + 3y)

9x² + y² – 6xy – 54x – 62y + 241 = 0

Linked Question 2

The coordinates of A and B are

$(- \frac{8}{5}, \frac{2 \sqrt{161}}{15}) and (- \frac{9}{5}, \frac{8}{5})$

(3. 0) and (0. 2)

$(- \frac{8}{5}, \frac{2 \sqrt{161}}{15}) and (0, 2)$

$(3, 0) and (- \frac{9}{5}, \frac{8}{5})$

Clearly A is (3, 0)

$y = mx \pm \sqrt{9 m^{2} + 4}$

$4 - 3 m = \pm \sqrt{9 m^{2} + 4}$

16 + 9m² – 24m = 9m² + 4

$12 = 24 m \Rightarrow m = \frac{1}{2}$

$y = \frac{1}{2} x + \sqrt{\frac{9}{4} + 4}$

2y = x + 5

Let Be be (x₁, y₁)

equation of tangent at B

$\begin{array}{l} \frac{{xx}_{1}}{9} + \frac{{yy}_{1}}{4} = 1 \\ x - 2 y = - 5 \end{array}} \Rightarrow \frac{x_{1}}{9 \cdot 1} = \frac{y_{1}}{4 (- 2)} = \frac{1}{- 5}$

$x_{1} = \frac{- 9}{5}, y_{1} = \frac{8}{5}$

Linked Question 3

The orthocenter of the triangle PAB is

$(\frac{8}{25}, \frac{7}{25})$

$(5, \frac{8}{7})$

$(\frac{11}{5}, \frac{8}{5})$

$(\frac{7}{5}, \frac{25}{8})$

$m_{AB} = \frac{\frac{8}{5} - 0}{\frac{- 9}{5} - 3} = \frac{8}{- 24} = \frac{- 1}{3}$

$∴$ Altitude P₂ : y – 4 = 3(x – 3)

$\Rightarrow$ 3x – y = 5 ....(1)

Altitude BM : y = $\frac{8}{5}$ ....(2)

Solving (1) and (2): $(\frac{11}{5}, \frac{8}{5})$