Question

Equation of tangents drawn from the point $(2,3)$ to the ellipse $9x^2+16y^2=144$ is:

• A. $y=4,x+y=3$
• B. $y=3,x+y=5$
• C. $y=3,x-y=5$
• D. $y+4=0,x-y=3$

Answer 1

Answer: (B)

$y=3,x+y=5$

Let $y=mx+c$ be a tangent to the curve.

Then it must also satisfy $9x^2+16y^2=144$.

Substituting in the equation of the curve, we get
$9x^2+16(mx+c{)}^2=144$

Now $y=mx+c$

It passes through $(2,3)$

Hence,
$3=2m+c$
Or
$c=3-2m$

Substituting above we get,
$9x^2+16(mx+c{)}^2=144$

$9x^2+16(mx+3-2m{)}^2=144$

$9x^2+16m^2{x}^2+32(3m-2m^2)\left(x\right)+16(3-2m{)}^2=144$.

If we substitute $x=0$, we get

$3-2m=9$

$c=9$

Then we get $m=0$ and $m=-1$

Hence, the equations are

$y=-x+5$ and $y=3$.