Question

Find the equations of tangents to the ellipse
$\frac{x^2}{25}+\frac{y^2}{16}=1$
which are parallel to $3x+2y=25$

• A. $3x+2y=17$
• B. $3x+2y+17=0$
• C. $3x+2y+32=0$
• D. $3x+2y=32$

Answer 1

Answer: (A), (B)

The correct options are
A

$3x+2y=17$

B

$3x+2y+17=0$

Any line parallel to $3x+2y=25$ can be written as $3x+2y=k$

We want to find the value of k. We can re - arrange
$3x+2y=k$ as $y=\frac{-3}{2}x+\frac{k}{2}$

Equation of tangent with Slope m to the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by $y=mx\mp \sqrt{a^2{m}^2+b^2}$

Here

$m=\frac{-3}{2}\Rightarrow \frac{k}{2}=\mp \sqrt{25\times {\left(\frac{-3}{2}\right)}^2+16}$

$\Rightarrow \frac{k}{2}=\mp \sqrt{{\left(\frac{5\times 3}{2}\right)}^2+16}$

$=\mp \sqrt{\frac{225+64}{4}}$

$=\mp \sqrt{\frac{289}{4}}$

$=\mp \frac{17}{2}$

$\Rightarrow k=\mp 17$

So the equation of tangents are $3x+2y=17$and $3x+2y+17=0$