Question

Tangents are drawn from the point $ P\left( 3,4 \right) $ to the ellipse $ \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1 $ touching the ellipse at points A and B. The coordinates of A and B are
A. $ \left( 3,0 \right) $ and $ \left( 0,2 \right) $
B. $ \left( -8,\dfrac{2\sqrt{161}}{15} \right) $ and $ \left( -\dfrac{9}{5},\dfrac{8}{5} \right) $
C. $ \left( -8,\dfrac{2\sqrt{161}}{15} \right) $ and $ \left( 0,2 \right) $
D. $ \left( 3,0 \right) $ and $ \left( -\dfrac{9}{5},\dfrac{8}{5} \right) $

Answer 1

Hint:
 We first describe the general condition of two tangents from a fixed point on an ellipse. We put the values for the point $ P\left( 3,4 \right) $ to the ellipse $ \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1 $ . We get the equations of the tangents. These lines touch the ellipse $ \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1 $ at points A and B. we find the intersecting points.

Complete step by step answer:
Tangents are drawn from the point $ P\left( 3,4 \right) $ to the ellipse $ \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1 $ touching the ellipse at points A and B.
From a fixed point $ \left( m,n \right) $ in general two tangents can be drawn to an ellipse. The equation of pair of tangents drawn to the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ is given by
 $ \left( \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}-1 \right)\left( \dfrac{{{m}^{2}}}{{{a}^{2}}}+\dfrac{{{n}^{2}}}{{{b}^{2}}}-1 \right)={{\left( \dfrac{mx}{{{a}^{2}}}+\dfrac{ny}{{{b}^{2}}}-1 \right)}^{2}} $ .
In symbol we write $ S{{S}_{1}}={{T}^{2}} $ , where \[S=\left( \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}-1 \right),{{S}_{1}}=\left( \dfrac{{{m}^{2}}}{{{a}^{2}}}+\dfrac{{{n}^{2}}}{{{b}^{2}}}-1 \right),T=\left( \dfrac{mx}{{{a}^{2}}}+\dfrac{ny}{{{b}^{2}}}-1 \right)\].
For our given ellipse $ \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1 $ and the point $ P\left( 3,4 \right) $ , the pair of tangents will be
 $ \begin{align}
  & \left( \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}-1 \right)\left( \dfrac{{{3}^{2}}}{9}+\dfrac{{{4}^{2}}}{4}-1 \right)={{\left( \dfrac{3x}{9}+\dfrac{4y}{4}-1 \right)}^{2}} \\
 & \Rightarrow 4\left( \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}-1 \right)={{\left( \dfrac{x}{3}+y-1 \right)}^{2}} \\
\end{align} $
We solve the equations to get the lines
 $ \begin{align}
  & 4\left( \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}-1 \right)={{\left( \dfrac{x}{3}+y-1 \right)}^{2}} \\
 & \Rightarrow \dfrac{4{{x}^{2}}}{9}+{{y}^{2}}-4=\dfrac{{{x}^{2}}}{9}+{{y}^{2}}+1+\dfrac{2xy}{3}-2y-\dfrac{2x}{3} \\
 & \Rightarrow \dfrac{3{{x}^{2}}}{9}-\dfrac{2xy}{3}+2y+\dfrac{2x}{3}-5=0 \\
 & \Rightarrow {{x}^{2}}-2xy+6y+2x-15=0 \\
 & \Rightarrow \left( x-3 \right)\left( x-2y+5 \right)=0 \\
\end{align} $
We have multiplication of two equations as 0.
So, the lines are $ \left( x-3 \right)=0;\left( x-2y+5 \right)=0 $ . These lines are the tangents.
These lines touch the ellipse $ \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1 $ at points A and B. We find the points.
For the line $ \left( x-3 \right)=0 $ , we have $ x=3 $ . Therefore, $ \dfrac{{{3}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1\Rightarrow y=0 $ .
So, one point is $ \left( 3,0 \right) $ .
For the line $ \left( x-2y+5 \right)=0 $ , we have $ x=2y-5 $ . Therefore,
 $ \begin{align}
  & \dfrac{{{\left( 2y-5 \right)}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1 \\
 & \Rightarrow 25{{y}^{2}}-80y+64=0 \\
 & \Rightarrow y=\dfrac{80\pm \sqrt{{{80}^{2}}-4\times 25\times 64}}{2\times 25}=\dfrac{8}{5} \\
\end{align} $ .
Putting value of y, we get $ x=\dfrac{2\times 8}{5}-5=-\dfrac{9}{5} $
So, other point is $ \left( -\dfrac{9}{5},\dfrac{8}{5} \right) $ .

The points are $ \left( 3,0 \right) $ and $ \left( -\dfrac{9}{5},\dfrac{8}{5} \right) $ . The correct option is D.

Note:
We need to remember that we can also solve this from the chord of contact points. We have the equation of chord of contact from the endpoints which are on the ellipse. We find the tangent equations from those points. Their intersecting point will be $ P\left( 3,4 \right) $ .