Question

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

Answer 1

Hint: - Equation of tangent in parametric form to the given ellipse is written as
$\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1$

Complete step-by-step answer:

Here we need to find the condition for the given line to be a tangent to the given ellipse ie.
So, we can proceed as we know that the equation of tangent in parametric form to the given ellipse is written as $\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1$.
Now as the next step compare the above parametric equation with \[x\cos \alpha + y\sin \alpha = p\]
This implies \[\dfrac{x}{p}\cos \alpha = \dfrac{x}{a}\cos \theta \] and \[\dfrac{y}{p}\sin \alpha = \dfrac{y}{b}\sin \theta ,\] if we rearrange these expressions, we get
\[\dfrac{{\cos \theta }}{{a\cos \alpha }} = \dfrac{{\sin \theta }}{{b\sin \alpha }} = \dfrac{1}{p}\]or \[ \Rightarrow \cos \theta = \dfrac{{a\cos \alpha }}{p}\]and \[{\text{sin}}\theta {\text{ = }}\dfrac{{b\sin \alpha }}{p}\]
We have obtained the value of \[\cos \theta \] and \[\sin \theta \]. Square and adding both sides we get
\[{\text{1 = }}\dfrac{{{a^2}{{\cos }^2}\alpha }}{{{p^2}}} + \dfrac{{{b^2}{{\sin }^2}\alpha }}{{{p^2}}}\]
\[ \Rightarrow {a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = {p^2}\]
Which is the required Condition for the given line to be a tangent on the given ellipse.

Note- Whenever this type of question appears always first note down the given details given in the question as it provides a better understanding to approach the question. Once \[\cos \theta \] and \[\sin \theta \] obtain square and add both sides respectively to obtain the required condition for the given line to be tangent on the given ellipse.