Question

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

• A. $3y=2x$
• B. $y-2x=0$
• C. $2x+y=3$
• D. None of the above

Answer 1

Answer: (A)

$3y=2x$

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

Equation of chord of contact is $T=0$

$x{x}_1+4y{y}_1=9$

$x\left(3\right)+4y\left(2\right)=9$

$3x+8y=9$ ......(i)

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

Equation of chord of contact when mid point is given is $T=S^{\prime }$

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

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

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

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

$\Rightarrow 2h=3k$ ......(iv)

Substituting (iv) in (iii), we get

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

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

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

Hence, proved.