Question

Find the condition for the line $x\cos \alpha +y\sin \alpha =p$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Answer 1

Step 1: Convert the line equation into slope-intercept form

Given line equation is$x\cos \alpha +y\sin \alpha =p$ and ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

The slope-intercept form of a line is given as,

$y=mx+c$

where, $m$ is the slope and $c$ is the intercept.

So,

$\begin{array}{rcl}& & \begin{array}{cc}& x\cos \alpha +y\sin \alpha =p\\ \Rightarrow & y\sin \alpha =-x\cos \alpha +p\\ \Rightarrow & y=-x\left(\frac{\cos \alpha }{\sin \alpha }\right)+\frac{p}{\sin \alpha }\end{array}\end{array}$

On comparing it with $y=mx+c$ we get,

$m=-\left(\frac{\cos \alpha }{\sin \alpha }\right)$ and $c=\frac{p}{\sin \alpha }$

Step 2: Evaluate the tangent condition

We know that,

If $y=mx+c$ is a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $c^2=a^2{m}^2+b^2$. Thus,

$\begin{array}{rcl}& & \begin{array}{cc}& {\left(\frac{p}{\sin \alpha }\right)}^2=a^2{\left(-\frac{\cos \alpha }{\sin \alpha }\right)}^2+b^2\\ \Rightarrow & \frac{p^2}{{\sin }^2\alpha }=a^2\frac{{\cos }^2\alpha }{{\sin }^2\alpha }+b^2\\ \Rightarrow & p^2=a^2{\cos }^2\alpha +b^2{\sin }^2\alpha \end{array}\end{array}$

Hence, the condition for the line $x\cos \alpha +y\sin \alpha =p$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is that $\begin{array}{rcl}{p}^2& =& a^2{\cos }^2\alpha +b^2{\sin }^2\alpha \end{array}$.