JEE Advance - Mathematics (2010 - Paper 2 Offline - No. 8)
Tangents are drawn from the point $$P(3, 4)$$ to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$ touching the ellipse at points $$A$$ and $$B$$.
The coordinates of $$A$$ and $$B$$ are
$$(3,0)$$ and $$(0,2)$$
$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$(0,2)$$
$$(3,0)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
Explanation
Equation of chord of contact is $${x \over 3} + y = 1$$, i.e. $$x = 3(1 - y)$$
Solving it with the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$
$${(1 - y)^2} + {{{y^2}} \over 4} = 1 \Rightarrow 4({y^2} - 2y + 1) + {y^2} = 4$$
$$\therefore$$ $$y = 0 \Rightarrow 5{y^2} - 8y = 0$$
$$\therefore$$ $$y = 0,\,8/5$$
Correspondingly $$x = 3,\, - 9/5$$
Points are $$(3,0)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
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