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Question

Tangents are drawn from the point $$\left(3, 2\right)$$ to the ellipse $${x}^{2}+4{y}^{2}=9$$. Find the equation to their chord of contact and the equation of the straight line joining $$\left(3, 2\right)$$ to the middle point of this chord of contact.


A
3y=2x
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B
y2x=0
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C
2x+y=3
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D
None of the above
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Solution

The correct option is C $$3y=2x$$

Given equation of ellipse is $${ x }^{ 2 }+4{ y }^{ 2 }=9$$

Equation of chord of contact is $$T=0$$

$$x{ x }_{ 1 }+4y{ y }_{ 1 }=9$$ 

$$x(3)+4y(2)=9$$ 

$$3x+8y=9$$    ......(i)

Let the middle point of chord of contact be $$(h,k)$$

Equation of chord of contact when mid point is given is $$T=S'$$

$$hx+4ky={ h }^{ 2 }+{ k }^{ 2 }$$    .....(ii)

Now (i) and (ii) are equation of same line 

$$\dfrac { h }{ 3 } =\dfrac { 4k }{ 8 } =\dfrac { { h }^{ 2 }+{ k }^{ 2 } }{ 9 } $$

$$ \Rightarrow 3h={ h }^{ 2 }+{ k }^{ 2 }$$    ......(iii)

$$ \Rightarrow 2h=3k$$    ......(iv)

Substituting (iv) in (iii), we get

$$27h=9{ h }^{ 2 }+4{ h }^{ 2 }\\ \Rightarrow 13{ h }^{ 2 }-27h=0\\ \Rightarrow h=0,\dfrac { 27 }{ 13 } $$

If $$h=0$$, then $$ k=0$$ 

So, the middle point of chord of contact is $$(0,0)$$

Equation of line joining $$(0,0)$$ and $$(3,2)$$ is

$$y-0=\dfrac { 2-0 }{ 3-0 } (x-0)\\ \Rightarrow 3y=2x$$

Hence, proved.


Maths

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