Question

$l_1$ is the tangent to $2x^2+3y^2=35$ at $(4,-1)$ and $l_2$ is the tangent to $4x^2+y^2=25$ at $(2,-3)$. The distance between $l_1$ and $l_2$ is:

• A. $\frac{10}{\sqrt{73}}$
• B. $\frac{60}{\sqrt{73}}$
• C. $0$
• D. $\frac{5}{3\sqrt{2}}$

Answer 1

Answer: (A)

$\frac{10}{\sqrt{73}}$

For the ellipse to be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

tangent at the point $(x_1,y_1)$ on it is given by $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

Equation of $l_1$ is $2x\left(4\right)+3y(-1)=35$
$\Rightarrow 8x-3y=35$
Equation of $l_2$ is $4x\left(2\right)+y(-3)=25$
$8x-3y=25$
$\therefore$ distance between $l_1$ and $l_2$ is $\frac{10}{\sqrt{(8{)}^2+(3{)}^2}}$
$=\frac{10}{\sqrt{73}}$