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 coordinates of A and B are [JEE Adv 2010 P2]

Correct Answer is : (3 , 0) and (−95,85)≤ft(-9/5,8/5)(−59,58).

≤ft(-8/5,{2{161}}{15}) and ≤ft(-9/8,8/5).

≤ft(-8/5,{2{161}}{15}) and (0 , 2).

Correct Answer is : (3 , 0) and (−95,85)≤ft(-9/5,8/5)(−59,58).

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Section: Mathematics

Question : 17 of 25

Marks: +1, -0

\frac{x^2}{9}+\frac{y^2}{4}

(3 , 0) and (0 , 2).
$\left(-\frac{8}{5},\frac{2\sqrt{161}}{15}\right)$ and $\left(-\frac{9}{8},\frac{8}{5}\right)$ .
$\left(-\frac{8}{5},\frac{2\sqrt{161}}{15}\right)$ and (0 , 2).
(3 , 0) and $\left(-\frac{9}{5},\frac{8}{5}\right)$ .

Solution:

We have

y = mx +

\sqrt{9m^2+4}

4 - 3m =

\sqrt{9m^2+4}

16+9m^2-24m

9m^2+4

⇒ m =

\frac{12}{24}

\frac{1}{2}

The equation is

y - 4 =

\frac{1}{2}(x-3)

2y - 8 = x - 3 ⇒ x - 2y + 5 = 0

Let B = (a, β).

⇒

\frac{x\alpha}{9}+\frac{y\beta}{4}-1

= 0 ⇒

\frac{\frac{\alpha}{9}}{1}

\frac{\frac{\beta}{4}}{-2}

\frac{-1}{5}

⇒ α =

\frac{-9}{5}

, β =

\frac{8}{5}

Hence, B =

\left(-\frac{9}{5},\frac{8}{5}\right)

. From the figure depicted here, the coordinates of point A is given by A ≡ (3, 0).

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